Exploring Algebraic Identities (Algebraic Identities ko Explore Karo): Class 9 Maths Chapter 4 Hinglish mein

Squares, strips, tiles aur cubes jo algebra samjhate hain

Ganita Manjari Part 1 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

107 min
30 practice questions
31 MCQs

Is chapter mein aap kya seekhenge

Identity aur equation mein fark pehchano, kisi claimed identity ko numbers se test karo aur distributive law se prove karo
(a + b)², (a − b)², (a + b + c)² aur a² − b² ko area models se samjhao, aur inse expressions expand karo aur numbers ka square mann mein nikalo
Perfect-square trinomials factorise karo, unko bhi jinmein common factor ya fractional coefficients hon
Quadratics ko algebra tiles se aur middle term split karke factorise karo, aur linear binomials ko jaldi multiply karo
Cube identities (a ± b)³, x³ ± y³ aur x³ + y³ + z³ − 3xyz derive karo aur use karo
Non-zero common factors cancel karke rational expressions simplify karo, aur aise simple problems solve karo jo quadratics tak le jaate hain

NCERT Mathematics Class 9, Chapter 4, pp. 68–91 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

4.1 Introduction

Yeh trick kisi dost par try karo. Usse lagaatar teen whole numbers chunne ko kaho, jaise 11, 12 aur 13, aur tumhe sirf beech wala number bataye. Phir bolo: “Sabse bade aur sabse chhote ka square karo, subtract karo, aur tumhe 48 milega.” Yeh har baar kaam karta hai, kyunki jawab hamesha beech wale number ka chaar guna hota hai: $13^{2} - 11^{2} = 169 - 121 = 48 = 4 \times 12$ .

Example 1NCERT Example 1 · p. 68

Squares mein chhupa ek pattern

Teen consecutive numbers lo. Sabse bade aur sabse chhote ka square karo, aur subtract karo. Yeh 6, 7, 8 ke liye karo, phir 19, 20, 21 ke liye, aur phir 49, 50, 51 ke liye. Har jawab ko beech wale number se compare karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Teen try, teen baar match. Fig. 4.1 dikhata hai ki 5, 6 aur 7 ke liye aisa kyun hota hai. Tiles ke chhote square ko bade square par rakho: jo tiles dhaki nahi rehti, unhe 6 ki 4 rows mein arrange kiya ja sakta hai, aur 6 hi beech wala number hai.

4/4Strip ko beech se fold karo: 6 ki 4 rows, aur 6 beech wala number hai.

Fig. 4.1Jab

5^{2}

7^{2}

par rakhte hain, to bacha hua border 6 ki 4 rows mein arrange ho jaata hai.

Step 1 mein blue tiles ka 7 by 7 square hai jis par 49 likha hai, aur alag se green tiles ka 5 by 5 square hai jis par 25 likha hai. Step 2 mein green square ko blue square ke bottom-left corner mein rakha jaata hai; jo 24 tiles dhaki nahi hain, yaani upar ki do rows aur right ke do columns, woh orange ho jaati hain. Step 3 mein orange border utha liya jaata hai, peeche ek dashed outline bachti hai, aur border ko 2 tiles oonchi aur 12 tiles lambi strip ki tarah bichha diya jaata hai, kyunki 12 = 5 + 7. Step 4 mein strip fold hokar 6 tiles ki 4 rows ka block ban jaati hai, jis par likha hai ki 7 ka square minus 5 ka square barabar 4 guna 6, jahan 6 beech wala number hai.

Phir bhi, teen examples aur ek picture yeh prove nahi karte ki pattern teen numbers ke har set ke liye kaam karta hai, jaise 1000, 1001 aur 1002, ya $- 3$ , $- 2$ aur $- 1$ . Pakka hone ke liye hamein algebra chahiye, aur khaas taur par woh special equalities jinhe identities (सर्वसमिका) kehte hain. Hum is pattern ko Section 4.3 mein prove karenge.

Yeh chapter identities ka ek chhota toolkit banata hai. Har identity ke saath squares, rectangles ya cubes se bani ek picture hai, aur ek chhota algebraic proof. Tum inka use numbers ka square mann mein nikalne, expressions ko factors mein todne, aur polynomials se bane fractions simplify karne ke liye karoge.

Quick check

3 mein se 1

Pattern ka use karke bina square kiye

31^{2} - 29^{2}

nikalo.

Answer dekho

Answer 120

Teen numbers 29, 30, 31 hain, isliye jawab

4 \times 30 = 120

hai. Check:

961 - 841 = 120

4.2 Identities ko Visualise Karna

Ek square garden plot socho jiski side do hisson se bani hai, ek lamba hissa aur ek chhota. Plot apne aap kyariyon mein bat jaata hai, aur kyariyon ko jodna asal mein chhupa hua algebra hai. Jab letters lengths ke liye hon, to $a b$ ek rectangle ka area hai aur $a^{2}$ ek square ka area. Tab identity shapes ko kaatne aur wapas jodne ke baare mein ek claim ban jaati hai.

Fig. 4.2 aisa square ek-ek piece karke banata hai. Shuru mein side $a$ ka square hai, phir use $b$ se chauda karta hai, $b$ se ooncha karta hai, aur aakhir mein corner mein bacha gap bhar deta hai. End tak har side $a + b$ ki ho jaati hai.

6/7Side $a + b$ hai, isliye $(a + b)^{2} = a^{2} + 2 a b + b^{2}$ .

Fig. 4.2Same area ko ek square ki tarah aur chaar pieces ki tarah ginne se

(a + b)^{2} = a^{2} + 2 a b + b^{2}

milta hai.NCERT Fig. 4.2 · p. 69

Step 1 mein length a ka blue segment aur length b ka green segment jud kar length a plus b ki line banate hain. Step 2 mein side a ka blue square banta hai, area a squared. Step 3 mein width b aur area ab ki orange strip uske right edge se lag jaati hai. Step 4 mein area ab ki doosri orange strip uske top edge par aa jaati hai. Step 5 mein top-right corner mein area b squared ka chhota green square ban jaata hai. Step 6 poore side a plus b wale square ko outline karta hai aur batata hai ki uska area a squared plus 2ab plus b squared hai. Step 7 dono strips hata deta hai, wahan question marks wale dashed holes bachte hain, yeh dikhane ke liye ki a squared plus b squared mein 2ab chhoot jaata hai.

Bade square ka area do tarah se gina ja sakta hai. Ek square ki tarah yeh $(a + b)^{2}$ hai. Piece by piece yeh $a^{2} + a b + a b + b^{2}$ hai. Ek hi area do baar ginne par jawab ek hi aana chahiye, isliye

Square of a sum

(a + b)^{2} = a^{2} + 2 a b + b^{2}

Ab formula ko todne ki koshish karo. Ise aisi values do jo koi drawing handle nahi kar sakti: ek negative number, fractions ka ek pair, decimals ka ek pair.Square ki side kabhi negative nahi hoti, isliye picture $a = - 5$ jaisi value ki guarantee nahi de sakti. Agar formula in sab mein bhi sahi nikle, to hamare paas lengths ki duniya ke bahar bhi us par bharosa karne ki achhi wajah hai.

Example 2NCERT Example 2 · p. 70

Lengths ke bahar testing

Ek chhota test lab set up karo. Check karo ki $(a + b)^{2}$ aur $a^{2} + 2 a b + b^{2}$ barabar aate hain ya nahi jab (i) $a = - 5$ , $b = 2$ ; (ii) $a = \frac{1}{2}$ , $b = - \frac{1}{3}$ ; (iii) $a = 0.3$ , $b = 0.7$ .

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Formula har test mein paas ho gaya. Lekin hum kitne bhi pairs try kar lein, anginat pairs bina check kiye reh jaate hain: ek peti se kuch aam check karne se yeh guarantee nahi milti ki baaki mein se koi daagi nahi hai. Hamein aisa argument chahiye jo ek saath har $a$ aur har $b$ ko cover kare, aur distributive law yahi deta hai.

TheoremSquare of a sum

Saare numbers $a$ aur $b$ ke liye, $(a + b)^{2} = a^{2} + 2 a b + b^{2}$ .

Definition

Identity

Do algebraic expressions ke beech aisi equality jo sahi rehti hai, chahe uske letters ko koi bhi values do.

Example $x (x + 3) = x^{2} + 3 x$ ek identity hai: $x = 2$ , $x = - 1$ ya $x = \frac{1}{2}$ try karo.

Ise ek normal equation (समीकरण) se compare karo. Equation $3 x - 5 = 16$ sirf tab sahi hai jab $x = 7$ ho; $x$ ki har doosri value ise galat bana deti hai. Equation solve karne ka matlab hai woh kuch values dhoondhna jo use sahi banati hain. Identity ko solve nahi karna padta, kyunki har value kaam karti hai.

Compare

Equation ya identity?

Equation

Sirf kuch khaas values ke liye sahi: $x^{2} + x = 6$ sirf $x = 2$ aur $x = - 3$ ke liye sahi hai
Hum un values ko dhoondhne ke liye ise solve karte hain
Example: $2 x + 1 = 11$ (sirf $x = 5$ )

Identity

Har letter ki har value ke liye sahi
Hum ise expand, factorise ya simplify karne ke liye use karte hain
Example: $2 (x + 3) = 2 x + 6$

Ek bahut aam galti hai $(a + b)^{2} = a^{2} + b^{2}$ likhna. Fig. 4.2 ka last step dikhata hai ki galti kahan hai: $a^{2} + b^{2}$ mein dono strips chhoot jaati hain aur isliye $2 a b$ gayab ho jaata hai. Jaise $a = 9$ aur $b = 4$ ke saath, $(a + b)^{2} = 169$ hai jabki $a^{2} + b^{2} = 97$ , yaani theek $2 \times 9 \times 4 = 72$ kam.

Identity expand karne ki ek recipe hai. $a$ aur $b$ ki jagah jo bhi ho (ek number, ek letter ya poora term), recipe wahi rehti hai. Pehle term ka square karo, dono terms ke product ka do guna jodo, phir doosre term ka square jodo.

Example 3NCERT Example 3 · p. 71

Binomial square expand karna

$(6 u + 5 v)^{2}$ expand karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example

Brackets ke andar fractions

${(\frac{2}{3} p + \frac{3}{2} q)}^{2}$ expand karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Yahi identity mushkil squares ko mental arithmetic mein badal deti hai. Number ko ek round hisse aur ek chhote hisse mein todo, aur teen aasaan pieces jodo.

Example 4NCERT Example 4 · p. 71

Square courtyard ka area

Ek square mandir courtyard ki sides 62 m hain. Lambi multiplication ke bina iska area nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

InteractiveArea model lab

Chaar tabs wala ek interactive area model: (a + b) squared, (a minus b) squared, a squared minus b squared, aur (a + b + c) squared. Sliders aur plus/minus buttons se a ko 1 se 60, b ko 0 se 30 aur c ko 0 se 20 tak set kar sakte ho. Square scale ke hisaab se dobara banta hai, uske pieces rangeen hote hain aur un par unke areas likhe hote hain, aur neeche ek panel chune gaye numbers se identity ki dono sides nikalta hai. a squared minus b squared wale tab mein ek Rearrange button bache hue L-shape ko sides a + b aur a minus b wale rectangle mein badal deta hai. Panel ke neeche ek note b = 0 ya a = b jaise special cases samjhata hai.

Fig. 4.3Ek identity chuno, phir

a

b

(aur

c

) badlo: pieces, unke areas aur identity ki dono sides saath-saath update hoti hain.

Practice · 3 questions

Practice 4.1: answers ke saath

1
Short answerEasyNCERT Exercise Set 4.1, Q1
$(a + b)^{2} = a^{2} + 2 a b + b^{2}$ use karke expand karo.
1. $(8 a + 3 b)^{2}$
  Solution dikhao
  Answer $64 a^{2} + 48 a b + 9 b^{2}$
  $(8 a)^{2} + 2 (8 a) (3 b) + (3 b)^{2} = 64 a^{2} + 48 a b + 9 b^{2}$ .
2. ${(\frac{4}{5} m + \frac{5}{2} n)}^{2}$
  Solution dikhao
  Answer $\frac{16}{25} m^{2} + 4 m n + \frac{25}{4} n^{2}$
  Middle term $2 \times \frac{4}{5} \times \frac{5}{2} m n = 4 m n$ hai, isliye jawab $\frac{16}{25} m^{2} + 4 m n + \frac{25}{4} n^{2}$ hai.
3. $(1.2 x + 0.5 y)^{2}$
  Solution dikhao
  Answer $1.44 x^{2} + 1.2 x y + 0.25 y^{2}$
  ${1.2}^{2} = 1.44$ , $2 \times 1.2 \times 0.5 = 1.2$ aur ${0.5}^{2} = 0.25$ , to jawab $1.44 x^{2} + 1.2 x y + 0.25 y^{2}$ hai.
4. ${(\frac{2}{3} k + 9)}^{2}$
  Solution dikhao
  Answer $\frac{4}{9} k^{2} + 12 k + 81$
  ${(\frac{2}{3} k)}^{2} + 2 \cdot \frac{2}{3} k \cdot 9 + 81 = \frac{4}{9} k^{2} + 12 k + 81$ .
5. ${(y + \frac{1}{3 z})}^{2}$
  Solution dikhao
  Answer $y^{2} + \frac{2 y}{3 z} + \frac{1}{9 z^{2}}$
  $y^{2} + 2 \cdot y \cdot \frac{1}{3 z} + \frac{1}{9 z^{2}} = y^{2} + \frac{2 y}{3 z} + \frac{1}{9 z^{2}}$ .
6. ${(\frac{1}{p} + \frac{2}{q})}^{2}$
  Solution dikhao
  Answer $\frac{1}{p^{2}} + \frac{4}{p q} + \frac{4}{q^{2}}$
  $\frac{1}{p^{2}} + 2 \cdot \frac{1}{p} \cdot \frac{2}{q} + \frac{4}{q^{2}} = \frac{1}{p^{2}} + \frac{4}{p q} + \frac{4}{q^{2}}$ .
2
NumericalEasyNCERT Exercise Set 4.1, Q2
Isi identity se har square nikalo.
1. $73^{2}$
  Answer dekho
  Answer 5329
  $(70 + 3)^{2} = 4900 + 420 + 9 = 5329$ .
2. $102^{2}$
  Answer dekho
  Answer 10404
  $(100 + 2)^{2} = 10000 + 400 + 4 = 10404$ .
3. $306^{2}$
  Answer dekho
  Answer 93636
  $(300 + 6)^{2} = 90000 + 3600 + 36 = 93636$ .
Quick answerEasy
Inmein se kaunsa identity hai?
$x^{2} = 4 x$ $(y + 2)^{2} = y^{2} + 4$ $(m + n)^{2} = m^{2} + 2 m n + n^{2}$ $3 t = t + 6$
Answer dekho
Answer (C) $(m + n)^{2} = m^{2} + 2 m n + n^{2}$
Sirf teesra letters ki har value ke liye sahi hai. Pehla aur last sirf kuch khaas values ke liye sahi hain ( $x = 0$ ya $x = 4$ ; $t = 3$ ), aur doosra tab galat hai jab bhi $y \neq 0$ ho.

4.3 Identities ki madad se Algebraic Expressions ka Factorisation

Identity ko dono directions mein padha ja sakta hai. Left se right, $(a + b)^{2} = a^{2} + 2 a b + b^{2}$ ek square ko expand karti hai. Right se left, yeh kehti hai ki $a^{2} + 2 a b + b^{2}$ jaise shape wala expression asal mein $a + b$ ka square hai. Kisi expression ko product ki tarah dobara likhna factorisation (गुणनखंडन) kehlata hai, aur product ka har hissa ek factor hai.

Hum perfect-square trinomial (पूर्ण वर्ग) kaise pehchanein? Teen checks karo. Kya pehla term ek square hai? Kya last term ek square hai? Kya middle term unke square roots ke product ka theek do guna hai? Fig. 4.4 yeh checks $49 m^{2} + 42 m + 9$ par chalata hai, aur phir ek milte-julte expression par jo fail ho jaata hai.

4/5Isliye $49 m^{2} + 42 m + 9 = (7 m + 3)^{2}$ .

Fig. 4.4Trinomial perfect square tabhi hai jab uske end terms squares hon aur middle term unke roots ke product ka do guna ho.

Step 1 mein expression 49m squared plus 42m plus 9 dikhaya gaya hai. Step 2 pehle aur last terms ko box karta hai aur note karta hai ki woh 7m aur 3 ke squares hain. Step 3 middle term ko box karta hai aur check karta hai ki 2 times 7m times 3 barabar 42m hai, saath mein ek tick. Step 4 result likhta hai: 49m squared plus 42m plus 9 barabar (7m + 3) squared. Step 5 milta-julta expression 49m squared plus 40m plus 9 dikhata hai, jiska middle term 42m hona chahiye tha lekin 40m hai, isliye yeh perfect square nahi hai.

Example 5NCERT Example 5 · p. 72

Sabse simple case

$y^{2} + 10 y + 25$ factorise karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 6NCERT Example 6 · p. 72

Aage ek coefficient

$16 t^{2} + 56 t + 49$ factorise karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 7NCERT Example 7 · p. 72

Pehle common factor bahar nikalo

$3 x^{3} + 24 x^{2} + 48 x$ factorise karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example

Fractions se koi dikkat nahi

$4 x^{2} + 2 x + \frac{1}{4}$ factorise karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

TheoremSquare of a difference

Saare numbers $a$ aur $b$ ke liye, $(a - b)^{2} = a^{2} - 2 a b + b^{2}$ .

Ek area picture isko confirm karti hai (Fig. 4.5). Side $a$ ke square se shuru karo aur width $b$ ki do strips kaat do: ek side ke neeche tak aur ek top par aar-paar. Dono strips ek chhote $b \times b$ corner mein overlap karti hain, isliye woh corner do baar hat jaata hai. Use ek baar wapas dene par theek side $a - b$ ka square bachta hai.

4/5Corner square $b^{2}$ ko ek baar wapas rakho: $a^{2} - 2 a b + b^{2}$ .

Fig. 4.5Area

a b

ki do strips hat jaati hain, lekin unka common corner do baar gina gaya tha, isliye

b^{2}

wapas joda jaata hai.NCERT Fig. 4.3 · p. 74

Step 1 mein side a ka blue square hai, area a squared. Step 2 mein width b ki ek strip uski right side se khisak kar alag hoti hai; strip orange ho jaati hai aur us par ab likha hota hai, aur a squared minus ab bachta hai. Step 3 top strip ko, jiska area bhi ab hai, orange mein mark karta hai; uska right end us corner ke upar hai jo pehle hi hat chuka hai, jise hatched dikhaya gaya hai, isliye corner do baar hata hai. Step 4 us corner mein green square b squared wapas rakhta hai, jisse a squared minus 2ab plus b squared milta hai. Step 5 sirf bacha hua blue square dikhata hai, jiski sides a minus b hain, aur identity (a minus b) squared barabar a squared minus 2ab plus b squared.

Ab jab dono square identities hamare paas hain, to chapter ki shuruaat wali trick ke liye bas algebra ki ek line chahiye. Sab kuch beech wale number $n$ ke terms mein likha gaya hai.

TheoremPattern hamesha kyun kaam karta hai

Har number $n$ ke liye, $(n + 1)^{2} - (n - 1)^{2} = 4 n$ .

Square of a difference un numbers ke liye kaam ka hai jo kisi round number se thoda kam hain.

Example 8NCERT Example 8 · p. 73

Sau se thoda kam

$98^{2}$ mann mein nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Practice · 2 questions

Practice 4.2: answers ke saath

1
Short answerMediumNCERT Exercise Set 4.2, Q1
Poori tarah factorise karo.
1. $25 x^{2} + 30 x y + 9 y^{2}$
  Solution dikhao
  Answer $(5 x + 3 y)^{2}$
  $(5 x)^{2} + 2 (5 x) (3 y) + (3 y)^{2} = (5 x + 3 y)^{2}$ .
2. $16 m^{2} + 8 m n + n^{2}$
  Solution dikhao
  Answer $(4 m + n)^{2}$
  $16 m^{2} = (4 m)^{2}$ aur $2 \times 4 m \times n = 8 m n$ , isliye $(4 m + n)^{2}$ .
3. $81 p^{2} + 36 p q + 4 q^{2}$
  Solution dikhao
  Answer $(9 p + 2 q)^{2}$
  $(9 p)^{2} + 2 (9 p) (2 q) + (2 q)^{2} = (9 p + 2 q)^{2}$ .
4. $36 a^{2} + 16 a b + \frac{16}{9} b^{2}$
  Solution dikhao
  Answer ${(6 a + \frac{4}{3} b)}^{2}$
  $\frac{16}{9} b^{2} = {(\frac{4}{3} b)}^{2}$ aur $2 \times 6 a \times \frac{4}{3} b = 16 a b$ , isliye jawab ${(6 a + \frac{4}{3} b)}^{2}$ hai.
5. $5 x^{2} + 20 x y + 20 y^{2}$
  Solution dikhao
  Answer $5 (x + 2 y)^{2}$
  $5 x^{2} + 20 x y + 20 y^{2} = 5 (x^{2} + 4 x y + 4 y^{2}) = 5 (x + 2 y)^{2}$ .
6. $\frac{2}{3} p^{2} + 4 p q + 6 q^{2}$
  Solution dikhao
  Answer $\frac{2}{3} (p + 3 q)^{2}$
  $\frac{2}{3}$ bahar nikalo: $\frac{2}{3} (p^{2} + 6 p q + 9 q^{2}) = \frac{2}{3} (p + 3 q)^{2}$ . Check: $\frac{2}{3} \times 6 = 4$ aur $\frac{2}{3} \times 9 = 6$ .
2
NumericalEasyNCERT Exercise Set 4.2, Q2
$(a - b)^{2} = a^{2} - 2 a b + b^{2}$ use karke har square nikalo.
1. $88^{2}$
  Answer dekho
  Answer 7744
  $(90 - 2)^{2} = 8100 - 360 + 4 = 7744$ .
2. $197^{2}$
  Answer dekho
  Answer 38809
  $(200 - 3)^{2} = 40000 - 1200 + 9 = 38809$ .
3. $495^{2}$
  Answer dekho
  Answer 245025
  $(500 - 5)^{2} = 250000 - 5000 + 25 = 245025$ .

4.4 Aur Identities

Do terms ke sum ke squares ab jaane-pehchane hain. $(a + b + c)^{2}$ jaise teen terms ke sum ke liye terms ke saare nau pairs multiply karna kaam karega, lekin ek shortcut hai: $b + c$ ko ek number maan lo aur do terms wali identity dobara use karo.

DerivationSquare of a three-term sum

Maan lo $d = b + c$ . Tab $a + b + c = a + d$ , aur square of a sum se milta hai

(a + b + c)^{2} = (a + d)^{2} = a^{2} + 2 a d + d^{2}

Ab $d = b + c$ wapas rakho aur har piece expand karo:

= a^{2} + 2 a (b + c) + (b + c)^{2} = a^{2} + 2 a b + 2 a c + b^{2} + 2 b c + c^{2}

Aise group kiya ki yaad rakhna aasaan ho:

Square of a three-term sum

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a

Shabdon mein: teeno terms mein se har ek ka square karo, phir har pair ke product ka do guna jodo. Teen terms se teen pairs bante hain, $a b$ , $b c$ aur $c a$ . Fig. 4.6 yahi substitution areas ke saath karta hai.

3/3Teen squares aur barabar rectangles ke teen pairs: $2 a b$ , $2 b c$ aur $2 c a$ .

Fig. 4.6

d = b + c

todne se chaar pieces nau ban jaate hain: teen squares aur teen matching pairs.NCERT Fig. 4.4 · p. 76

Step 1 mein side a plus d ka square chaar pieces mein bata hai: ek blue a squared, do orange pieces ad aur ek green d squared, aur sides par a aur d likha hai. Step 2 dashed lines banata hai jo har side d ko b aur c mein todti hain, to square ke nau pieces ho jaate hain jin par a squared, ab, ca, ab, b squared, bc, ca, bc aur c squared likha hai. Step 3 pieces ko naye rang deta hai: diagonal par teen squares a squared, b squared aur c squared, aur barabar pairs ab, bc aur ca ek jaise rangon mein, aur square ke bagal mein identity likhi hai.

Example 9NCERT Example 9 · p. 76

Teen-digit number ka square

213 ko teen hisson mein todkar $213^{2}$ nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Ab hamare toolkit mein teen identities hain:

$(a + b)^{2} = a^{2} + 2 a b + b^{2}$
$(a - b)^{2} = a^{2} - 2 a b + b^{2}$
$(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a$

Sahi identity chunna zaroori hai. $198^{2}$ ke liye $200 - 2$ wale split mein bas ek aasaan subtraction lagta hai, jabki $100 + 90 + 8$ mein chhah products lagte hain. Aisa split dhoondho jismein ek round number aur ek chhota bacha hua hissa ho.

Practice · 4 questions

Practice 4.3: answers ke saath

1
NumericalEasyNCERT Exercise Set 4.3, Q1
Har square nikalo. Pehle tay karo ki kaunsi identity kaam sabse aasaan banati hai.
1. $122^{2}$
  Answer dekho
  Answer 14884
  $(120 + 2)^{2} = 14400 + 480 + 4 = 14884$ .
2. $69^{2}$
  Answer dekho
  Answer 4761
  $(70 - 1)^{2} = 4900 - 140 + 1 = 4761$ .
3. $395^{2}$
  Answer dekho
  Answer 156025
  $(400 - 5)^{2} = 160000 - 4000 + 25 = 156025$ .
4. $312^{2}$
  Answer dekho
  Answer 97344
  $(300 + 10 + 2)^{2} = 90000 + 100 + 4 + 6000 + 40 + 1200 = 97344$ .
5. $1006^{2}$
  Answer dekho
  Answer 1012036
  $(1000 + 6)^{2} = 1000000 + 12000 + 36 = 1012036$ .
6. $2030^{2}$
  Answer dekho
  Answer 4120900
  $(2000 + 30)^{2} = 4000000 + 120000 + 900 = 4120900$ .
2
Short answerMediumNCERT Exercise Set 4.3, Q2
Sahi identity use karke factorise karo.
1. $25 k^{2} - 40 k + 16$
  Solution dikhao
  Answer $(5 k - 4)^{2}$
  $(5 k)^{2} - 2 (5 k) (4) + 4^{2} = (5 k - 4)^{2}$ .
2. $\frac{4}{9} a^{2} + 4 a b + 9 b^{2}$
  Solution dikhao
  Answer ${(\frac{2}{3} a + 3 b)}^{2}$
  $\frac{4}{9} a^{2} = {(\frac{2}{3} a)}^{2}$ , $9 b^{2} = (3 b)^{2}$ aur $2 \times \frac{2}{3} a \times 3 b = 4 a b$ .
3. $\frac{x^{2}}{4} + \frac{y^{2}}{9} + 4 z^{2} + \frac{x y}{3} - \frac{4 y z}{3} - 2 z x$
  Solution dikhao
  Answer ${(\frac{x}{2} + \frac{y}{3} - 2 z)}^{2}$
  Squares $\frac{x}{2}$ , $\frac{y}{3}$ aur $2 z$ se aate hain. Cross terms $- \frac{4 y z}{3}$ aur $- 2 z x$ negative hain jabki $+ \frac{x y}{3}$ positive hai, isliye minus sign $z$ -term ke saath jaata hai: $2 \cdot \frac{x}{2} \cdot \frac{y}{3} = \frac{x y}{3}$ , $2 \cdot \frac{y}{3} \cdot (- 2 z) = - \frac{4 y z}{3}$ aur $2 \cdot (- 2 z) \cdot \frac{x}{2} = - 2 z x$ .
4. $\frac{x^{2}}{9} + 2 + \frac{9}{x^{2}}$
  Solution dikhao
  Answer ${(\frac{x}{3} + \frac{3}{x})}^{2}$
  $2 \times \frac{x}{3} \times \frac{3}{x} = 2$ , isliye expression ${(\frac{x}{3} + \frac{3}{x})}^{2}$ hai.
5. $4 p^{2} + 9 q^{2} + r^{2} + 12 p q - 4 p r - 6 q r$
  Solution dikhao
  Answer $(2 p + 3 q - r)^{2}$
  Roots $2 p$ , $3 q$ , $r$ hain. $r$ wale dono cross terms negative hain, isliye $- r$ lo: $2 (2 p) (3 q) = 12 p q$ , $2 (3 q) (- r) = - 6 q r$ , $2 (- r) (2 p) = - 4 p r$ .
3
Short answerEasyNCERT Exercise Set 4.3, Q3
Square of a three-term sum use karke expand karo.
1. $(2 a + b + 5 c)^{2}$
  Solution dikhao
  Answer $4 a^{2} + b^{2} + 25 c^{2} + 4 a b + 10 b c + 20 c a$
  Squares: $4 a^{2}$ , $b^{2}$ , $25 c^{2}$ . Pairs: $2 (2 a) (b) = 4 a b$ , $2 (b) (5 c) = 10 b c$ , $2 (5 c) (2 a) = 20 c a$ .
2. $(4 m - 3 n - 2 k)^{2}$
  Solution dikhao
  Answer $16 m^{2} + 9 n^{2} + 4 k^{2} - 24 m n + 12 n k - 16 k m$
  Squares: $16 m^{2}$ , $9 n^{2}$ , $4 k^{2}$ . Pairs: $2 (4 m) (- 3 n) = - 24 m n$ , $2 (- 3 n) (- 2 k) = 12 n k$ , $2 (- 2 k) (4 m) = - 16 k m$ .
4
Short answerMediumNCERT Exercise Set 4.3, Q4
Kya har statement ek identity hai? Agar yeh letters ki saari values ke liye sahi hai to True answer karo.
1. $(a + b)^{2} + (a - b)^{2} = 2 a^{2} + 2 b^{2}$
  SahiGalat
  Answer dekho
  Answer Sahi
  Dono expansions jodne par $2 a b$ aur $- 2 a b$ cancel ho jaate hain, aur $2 a^{2} + 2 b^{2}$ bachta hai.
2. $(a + b + c)^{2} - (a - b - c)^{2} = 4 a b + 4 a c$
  SahiGalat
  Answer dekho
  Answer Sahi
  $d = b + c$ ke saath yeh $(a + d)^{2} - (a - d)^{2} = 4 a d = 4 a b + 4 a c$ hai.
3. $(a - b - c)^{2} = a^{2} - b^{2} - c^{2} - 2 a b - 2 b c - 2 c a$
  SahiGalat
  Answer dekho
  Answer Galat
  Teeno squared terms positive aate hain: $(a - b - c)^{2} = a^{2} + b^{2} + c^{2} - 2 a b + 2 b c - 2 c a$ . $a = 1$ , $b = 2$ , $c = 3$ ke saath left side 16 hai lekin right side $- 34$ .

Tum 47 ka square mann mein kaise nikaloge? Product $50 \times 44$ aasaan hai, aur yeh $47^{2}$ se sirf 9 kam hai. Iski wajah Class 8 wala difference of two squares hai, $a^{2} - b^{2} = (a + b) (a - b)$ , jismein $b^{2}$ ko doosri side le jaaya gaya hai:

a^{2} = (a + b) (a - b) + b^{2}

TheoremDifference of two squares

Saare numbers $a$ aur $b$ ke liye, $a^{2} - b^{2} = (a + b) (a - b)$ .

Fig. 4.7 yeh rearranged form areas ke saath dikhata hai. Side $a$ ke square se ek $b \times b$ corner kaat lo. Phir bache hue L-shape ko do strips mein kaato aur ek ko ghuma do, taaki dono milkar $(a + b)$ lamba aur $(a - b)$ ooncha rectangle bana dein.

4/5Patli strip ko ghumao aur chaudi strip ke bagal mein rakho: $(a + b)$ lamba aur $(a - b)$ ooncha rectangle.

Fig. 4.7Square

a^{2}

ek rectangle

(a + b) (a - b)

aur ek chhota square

b^{2}

ban jaata hai.NCERT Fig. 4.5 · p. 77

Step 1 mein side a ka blue square hai. Step 2 top-right se side b ka green corner square utha kar ek taraf rakh deta hai, aur area a squared minus b squared ka L-shape bachta hai. Step 3 chaude hisse ke top par ek cut mark karta hai, jo (a minus b) by b ki patli orange strip ko a by (a minus b) ki chaudi blue strip se alag karta hai. Step 4 patli strip ko right angle par ghuma kar chaudi strip ke right end se lagata hai, aur sides a plus b aur a minus b ka ek rectangle banta hai. Step 5 chhote square ko rectangle ke bagal mein rakhta hai aur batata hai ki a squared barabar (a + b)(a minus b) plus b squared.

Example

Dostana partners

$a^{2} = (a + b) (a - b) + b^{2}$ use karke (i) $96^{2}$ aur (ii) $38^{2}$ nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Socho zaraNCERT Think and Reflect · p. 78

Picture se identity padho

Chaar barabar rectangles, har ek $a$ by $b$ , ek chhote square ko gher kar ek bada square banate hain (Fig. 4.8). Bade square ka area do alag tareekon se nikalo. Tumhe kaunsi identity milti hai? Ise $a = 5$ aur $b = 2$ se check karo.

2/3Har rectangle $a$ by $b$ hai, aur beech wale square ki side $a - b$ hai.

Fig. 4.8Ek square ko do tarah se ginne par

(a + b)^{2} = (a - b)^{2} + 4 a b

milta hai.NCERT Fig. 4.6 · p. 78

Step 1 mein ek bada square hai jiski sides par a aur b mark hai; yeh chaar orange rectangles se bana hai jo ek chhote purple square ke chaaron taraf pinwheel ki tarah rakhe hain, aur poochha gaya hai ki yeh kaunsi identity dikhata hai. Step 2 chaaron rectangles ko thoda alag kheenchta hai aur har ek par ab likhta hai, aur beech ke square mein question mark hai, jiski side a minus b hai. Step 3 pieces ko wapas rakhta hai, beech ke square par (a minus b) squared likhta hai aur likhta hai (a + b) squared barabar (a minus b) squared plus 4ab.

Ek badi challenge: $(a + b + c)^{2} + (a + b - c)^{2} + (a - b + c)^{2} + (a - b - c)^{2}$ expand karo aur jodo. Lagbhag har term cancel ho jaata hai. Kya tum chhote jawab ko squares ke ek set ki tarah draw kar sakte ho?

4.5 Algebra Tiles se Factorisation

Algebra tiles multiplication aur factorisation ko ek aisi puzzle bana dete hain jise tum haath mein pakad sakte ho. Ek set mein teen tarah ke pieces hote hain: side $x$ ka bada square (area $x^{2}$ ), $x$ by 1 ki lambi strip (area $x$ ), aur 1 by 1 ka chhota square (area 1). Length $x$ jaan-boojh kar units ki whole number nahi rakhi gayi, isliye ek strip ko kabhi chhote squares ki row se badla nahi ja sakta.

$x^{2} + 6 x + 8$ jaisa expression tiles ka ek dher ban jaata hai: ek bada square, chhah strips aur aath chhote squares. Ise factorise karne ka matlab hai saari tiles ko bina gap aur bina overlap ke ek rectangle mein fit karna. Us rectangle ki do sides hi factors hain.

5/5Rectangle $x + 4$ chauda aur $x + 2$ ooncha hai, isliye $x^{2} + 6 x + 8 = (x + 2) (x + 4)$ .

Fig. 4.9Chhah strips ko 4 aur 2 mein baantne se 4 by 2 ka gap bachta hai, jo 8 unit tiles ke liye bilkul sahi hai.NCERT Fig. 4.7 · p. 79

Step 1 mein tiles ka ek dher hai: ek blue x-squared tile, chhah orange x-strips khadi hui aur aath chhoti green unit tiles. Step 2 x-squared tile ko right mein uski jagah par le jaata hai. Step 3 uske left mein chaar strips saath-saath khadi karta hai aur uske neeche do strips lita deta hai. Step 4 bottom-left corner ke 4 by 2 gap ko aath unit tiles se bilkul bhar deta hai; chaar unit tiles ki row side x se saaf lambi dikhti hai, kyunki x units ki whole number nahi hai. Step 5 width par x plus 4 aur height par x plus 2 likhta hai aur likhta hai x squared plus 6x plus 8 barabar (x + 2)(x + 4).

Yeh kaam kyun kiya? Bade square ke bagal wali strips aur uske neeche wali strips khaali corner ka size tay karti hain. Ek taraf 4 aur doosri taraf 2 hone se corner 4 by 2 ka hai, aur usmein dher ki theek 8 unit tiles aati hain. Strips ko kisi aur tarah baantne par ya to tiles bach jaati ya gap reh jaata. Bana hua rectangle ek saath do kahaniyan sunata hai:

uski sides multiply karne se uska area milta hai: $(x + 2) (x + 4) = x^{2} + 6 x + 8$ ;
tiles se uski sides padhne par area factorise ho jaata hai: $x^{2} + 6 x + 8 = (x + 2) (x + 4)$ .

InteractiveAlgebra tile board

Ek algebra tile board. Ek menu mein x squared + 6x + 8, x squared + 4x + 5 aur 2x squared + 5x + 2 jaise expressions hain. x-squared tiles corner mein rehti hain; plus aur minus buttons se tay hota hai ki right mein kitne unit columns aur neeche kitni rows lagengi. Board is shape ke liye zaroori strips aur unit tiles banata hai, tiles kam hon to khaali dashed jagah dikhata hai aur zyada hon to bacha hua dher, aur dono side lengths likhta hai. Cards ginte hain ki di gayi x-strips aur unit tiles mein se kitni use hui, aur rectangle poora hone par ek message factorisation batata hai. Ek button hint ke liye constant term ke factor pairs dikhata hai.

Fig. 4.10Ek expression chuno, phir unit columns aur rows jodte jao jab tak har tile ek rectangle mein fit na ho jaaye.

Try karke dekhoNCERT Think and Reflect · p. 79

Tile board par

Pehle guess karo

Banane se pehle: product

(x + 1) (x + 5)

ke liye kitni unit tiles chahiye?

56110

$x^{2} + 7 x + 10$ chuno aur rectangle banao. Iski sides kya hain?
$x^{2} + 9 x + 18$ chuno. Right mein pehle 1, phir 2, phir 3 unit columns try karo, har baar rows jodte jao jab tak saari 9 strips use na ho jaayein. Kaunsa choice saari 18 unit tiles bhi use kar leta hai?
$x^{2} + 4 x + 5$ chuno. Kya tum rectangle bana sakte ho? Jo milta hai use samjhane ke liye Show factor pairs dabao.

$x^{2} + 7 x + 10 = (x + 2) (x + 5)$ . $x^{2} + 9 x + 18$ ke liye columns aur rows ka sum 9 hona chahiye. 1 aur 8 se sirf 8 unit tiles use hoti hain, aur 2 aur 7 se sirf 14, lekin 3 aur 6 saari 18 use karte hain, jisse $(x + 3) (x + 6)$ milta hai. Chhah columns aur teen rows wahi rectangle ghuma kar banate hain. $x^{2} + 4 x + 5$ ke liye unit block sirf 1 by 5 ho sakta hai, jiske liye $1 + 5 = 6$ strips chahiye, lekin sirf 4 hain, isliye koi rectangle nahi banta.

$(x + 2) (x + 4) = x^{2} + 6 x + 8$ ke numbers ko tile-board wale rectangle $(x + 3) (x + 6) = x^{2} + 9 x + 18$ ke numbers se compare karo. Dono cases mein middle coefficient brackets ke do numbers ka sum hai, aur last term unka product. Brackets ko multiply karke dekho to pata chalta hai ki aisa hamesha kyun hota hai.

TheoremSame pehle term wale do binomials

Saare numbers $x$ , $a$ aur $b$ ke liye, $(x + a) (x + b) = x^{2} + (a + b) x + a b$ .

Tiles 1 se bade pehle coefficient ko bhi sambhal lete hain. $2 x^{2} + 5 x + 2$ factorise karne ke liye dono bade squares ko ek ke upar ek rakho. Fig. 4.11 dikhata hai ki paanch strips aur do unit tiles rectangle ko kaise poora karti hain.

4/4 $2 x^{2} + 5 x + 2 = (2 x + 1) (x + 2)$ ; strips ki ginti $1 + 2 \times 2 = 5$ hai.

Fig. 4.11

2 x^{2} + 5 x + 2

ki tiles

x + 2

chaude aur

2 x + 1

oonche rectangle mein fit hoti hain.NCERT Fig. 4.8 · p. 80

Step 1 mein tiles ek dher mein hain: do blue x-squared tiles, paanch orange strips aur do green unit tiles. Step 2 dono bade squares ko ek ke upar ek rakhta hai aur unke neeche ek strip lita deta hai, jisse 2x plus 1 oonchi column banti hai. Step 3 uske right mein do aur columns banata hai, har ek do khadi strips aur neeche ek unit tile se bani, isliye har ek bhi 2x plus 1 oonchi hai. Step 4 width par x plus 2 aur height par 2x plus 1 likhta hai aur likhta hai 2x squared plus 5x plus 2 barabar (2x + 1)(x + 2), aur note karta hai ki strips ki ginti 1 plus 2 times 2 barabar 5 hai.

5 kahan se aaya? Bade squares ke neeche ek strip, aur dono side columns mein do-do strips: $1 + 2 \times 2 = 5$ . Kinhi bhi do linear brackets ko multiply karne se general pattern dikhta hai.

TheoremDo general linear binomials

Saare numbers $p$ , $q$ , $a$ , $b$ aur $x$ ke liye, $(p x + a) (q x + b) = p q x^{2} + (p b + q a) x + a b$ .

4.6 Algebra Tiles ke Bina Factorisation

Bade numbers ke liye tiles thakane wali ho jaati hain: koi bhi $x^{2} + 25 x + 144$ ko tiles se nahi bichhana chahega. Achhi baat yeh hai ki tiles ne hamein pehle hi dikha diya hai ki kya dhoondhna hai. the product rule for two binomials ke hisaab se, $x^{2} + b x + c$ theek tab $(x + p) (x + q)$ mein factorise hota hai jab hum aise do numbers $p$ aur $q$ dhoondh sakein jinke liye

p + q = b aur p q = c

Product wali condition zyada sakht hai, isliye $c$ ke factor pairs ki list banao aur har pair ka sum check karo. Is method ko splitting the middle term kehte hain: jo pair kaam karta hai woh batata hai ki $b x$ ko $p x + q x$ mein kaise todna hai. Fig. 4.12 $x^{2} + 9 x + 20$ ke liye yeh khoj karta hai.

4/4Isliye $9 x = 4 x + 5 x$ aur $x^{2} + 9 x + 20 = (x + 4) (x + 5)$ .

Fig. 4.12Constant ke factor pairs mein woh pair dhoondho jiska sum middle coefficient ho.

Step 1 mein expression x squared plus 9x plus 20 aur goal dikhaya gaya hai: aise do numbers jinka product 20 aur sum 9 ho, aur neeche pair, product aur sum columns wali ek khaali table hai. Step 2 pairs 1 aur 20, sum 21, aur 2 aur 10, sum 12, bharta hai, aur dono par cross laga hai. Step 3 pair 4 aur 5, sum 9, jodta hai, jo tick ke saath highlight hai. Step 4 9x ko 4x plus 5x mein todta hai aur likhta hai x squared plus 9x plus 20 barabar (x + 4)(x + 5).

Example 10NCERT Example 10 · p. 81

Sum 9, product 20

$x^{2} + 9 x + 20$ factorise karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 11NCERT Example 11 · p. 81

Sahi product, galat sum

$x^{2} + 13 x + 36$ factorise karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 12NCERT Example 12 · p. 81

Negative middle term

$x^{2} - 9 x + 14$ factorise karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example

Negative constant term

$t^{2} + 2 t - 35$ factorise karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Table 4.1

x^{2} + b x + c

p

aur

q

ke signs padhna

$c$ hai	$b$ hai	$p$ aur $q$ hain	Example
positive	positive	dono positive	$x^{2} + 7 x + 6 = (x + 1) (x + 6)$
positive	negative	dono negative	$x^{2} - 8 x + 12 = (x - 2) (x - 6)$
negative	positive	ulte signs; 0 se zyada door wala positive	$x^{2} + 3 x - 10 = (x + 5) (x - 2)$
negative	negative	ulte signs; 0 se zyada door wala negative	$x^{2} - 3 x - 10 = (x - 5) (x + 2)$

Jab pehla coefficient 1 nahi hota, to ek badlaav chahiye. Pehle aur last coefficients ko multiply karo, phir aise do numbers dhoondho jinka product yeh ho aur sum middle coefficient ho. Iski wajah the general product se aati hai: middle coefficient ke do hisse, $p b$ aur $q a$ , multiply hokar $p q \times a b$ dete hain, yaani pehla coefficient times last.

Example

Pehla coefficient 3

$3 x^{2} + 10 x + 8$ factorise karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Practice · 3 questions

Practice 4.4: answers ke saath

1
Short answerMediumNCERT Exercise Set 4.4, Q1
Har identity poori karo.
1. $k^{2} - 13 k + 40 = (□) (□)$
  $(k - 5) (k - 8)$ $(k + 5) (k + 8)$ $(k - 4) (k - 10)$ $(k - 5) (k + 8)$
  Answer dekho
  Answer (A) $(k - 5) (k - 8)$
  Hamein product 40 aur sum $- 13$ chahiye: $- 5$ aur $- 8$ .
2. $(□) (x + 2) = 2 x^{2} + x - 6$
  $2 x - 3$ $2 x + 3$ $x - 3$ $2 x - 6$
  Answer dekho
  Answer (A) $2 x - 3$
  $(2 x - 3) (x + 2) = 2 x^{2} + 4 x - 3 x - 6 = 2 x^{2} + x - 6$ .
3. $6 x^{2} + □ x - 15 = (2 x + 3) (3 x - 5)$ . Box mein kaunsa number aayega?
  Answer dekho
  Answer -1
  $(2 x + 3) (3 x - 5) = 6 x^{2} - 10 x + 9 x - 15 = 6 x^{2} - x - 15$ , isliye box mein $- 1$ aayega.
4. $4 x^{2} + 8 x + 3 = (□) (□)$
  Solution dikhao
  Answer $(2 x + 1) (2 x + 3)$
  $4 \times 3 = 12$ ; product 12 aur sum 8 wale numbers 2 aur 6 hain. Isliye $4 x^{2} + 2 x + 6 x + 3 = 2 x (2 x + 1) + 3 (2 x + 1) = (2 x + 1) (2 x + 3)$ .
2
NumericalEasyNCERT Exercise Set 4.4, Q2
Sahi identity chuno aur seedha multiply kiye bina har value nikalo.
1. $52^{2}$
  Answer dekho
  Answer 2704
  $(50 + 2)^{2} = 2500 + 200 + 4 = 2704$ .
2. $39^{2}$
  Answer dekho
  Answer 1521
  $(40 - 1)^{2} = 1600 - 80 + 1 = 1521$ .
3. $26 \times 14$
  Answer dekho
  Answer 364
  $(20 + 6) (20 - 6) = 400 - 36 = 364$ .
4. $115^{2}$
  Answer dekho
  Answer 13225
  $(100 + 15)^{2} = 10000 + 3000 + 225 = 13225$ .
5. $94^{2}$
  Answer dekho
  Answer 8836
  $(100 - 6)^{2} = 10000 - 1200 + 36 = 8836$ .
6. $19 \times 26$
  Answer dekho
  Answer 494
  $(x + a) (x + b)$ use karke, $(20 - 1) (20 + 6) = 400 + (6 - 1) \times 20 - 6 = 494$ .
7. $37 \times 44$
  Answer dekho
  Answer 1628
  $(40 - 3) (40 + 4) = 1600 + 1 \times 40 - 12 = 1628$ .
8. $305^{2}$
  Answer dekho
  Answer 93025
  $(300 + 5)^{2} = 90000 + 3000 + 25 = 93025$ .
3
Short answerMediumNCERT Exercise Set 4.4, Q3
Factorise karo.
1. $4 x^{2} + y^{2} + 9 z^{2} + 4 x y - 12 x z - 6 y z$
  Solution dikhao
  Answer $(2 x + y - 3 z)^{2}$
  Roots $2 x$ , $y$ , $3 z$ hain; $z$ wale dono cross terms negative hain, isliye $- 3 z$ use karo: $2 (2 x) (y) = 4 x y$ , $2 (2 x) (- 3 z) = - 12 x z$ , $2 (y) (- 3 z) = - 6 y z$ .
2. $9 m^{2} + 16 n^{2} - 24 m n$
  Solution dikhao
  Answer $(3 m - 4 n)^{2}$
  $(3 m)^{2} - 2 (3 m) (4 n) + (4 n)^{2} = (3 m - 4 n)^{2}$ .
3. $t^{2} + 3 t - 40$
  Solution dikhao
  Answer $(t + 8) (t - 5)$
  Product $- 40$ , sum 3: 8 aur $- 5$ .
4. $25 p^{2} + 20 p q + 4 q^{2}$
  Solution dikhao
  Answer $(5 p + 2 q)^{2}$
  $2 (5 p) (2 q) = 20 p q$ , isliye $(5 p + 2 q)^{2}$ .
5. $36 a^{2} + 4 b^{2} + 49 c^{2} - 24 a b + 84 a c - 28 b c$
  Solution dikhao
  Answer $(6 a - 2 b + 7 c)^{2}$
  Roots $6 a$ , $2 b$ , $7 c$ hain. $b$ wale cross terms negative hain, isliye $- 2 b$ lo: $2 (6 a) (- 2 b) = - 24 a b$ , $2 (6 a) (7 c) = 84 a c$ , $2 (- 2 b) (7 c) = - 28 b c$ .

4.7 Nayi Identities Dhoondhna

Squares se hamein bahut kuch mila. Cubes ke saath kya hota hai? $(a + b)^{2}$ ko ek aur factor $a + b$ se multiply karo:

DerivationCube of a sum

(a + b)^{3} = (a + b) (a^{2} + 2 a b + b^{2})

= a^{3} + 2 a^{2} b + a b^{2} + a^{2} b + 2 a b^{2} + b^{3}

Like terms ikattha karne par:

Cube of a sum

(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}

Volume model mein edge $a + b$ ka ek cube liya jaata hai, jise har edge par wahan kaata gaya hai jahan $a$ aur $b$ milte hain (Fig. 4.13). Isse aath blocks bante hain. Unhe is hisaab se baanto ki unke teen edges mein se kitne $b$ lambe hain. Bina $b$ -edge wala ek block hai, $a^{3}$ . Theek ek aise edge wale teen hain, kyunki $b$ -edge teen directions mein ho sakta hai: $3 a^{2} b$ . Theek do aise edges wale bhi teen hain: $3 a b^{2}$ . Teen $b$ -edges wala ek block hai, $b^{3}$ .

6/6 $(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}$ .

Fig. 4.13Aath blocks milkar cube banate hain:

(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}

.NCERT Fig. 4.10 · p. 83

Step 1 mein teen dimensions mein bana ek cube hai jiska har edge ek length a aur ek length b se bana hai. Step 2 aathon blocks ko alag kheenchta hai: aage neeche ek bada blue cube, peeche upar ek chhota pink cube, teen orange slabs aur teen green rods. Step 3 dono cubes, a cubed aur b cubed, ko highlight karta hai. Step 4 teen orange slabs ko highlight karta hai, har ek a by a by b, jin par a squared b likha hai. Step 5 teen green rods ko highlight karta hai, har ek a by b by b, jin par a b squared likha hai. Step 6 aathon blocks ko identity ke saath dikhata hai: (a + b) cubed barabar a cubed plus 3a squared b plus 3a b squared plus b cubed.

TheoremCube of a difference

Saare numbers $a$ aur $b$ ke liye, $(a - b)^{3} = a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3}$ .

Example 13NCERT Example 13 · p. 84

Perfect cube pehchanna

Dikhao ki $27 a^{3} + 27 a^{2} b + 9 a b^{2} + b^{3}$ ek perfect cube hai. Phir iska use karke $31^{3}$ nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 14NCERT Example 14 · p. 84

Kya yeh cube of a difference hai?

Tay karo ki $64 x^{3} - 144 x^{2} y + 108 x y^{2} - 27 y^{3}$ kisi binomial ka cube hai ya nahi. Agar hai, to binomial nikalo aur numbers se check karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example

100 ke paas ke number ka cube

$102^{3}$ nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Kuch products hairaan kar dene wale tareeke se chhote ho jaate hain. $x - y$ ko $x^{2} + x y + y^{2}$ se ek grid ki madad se multiply karo, har cell mein har bracket se ek term rakh kar (Fig. 4.14).

4/4 $(x - y) (x^{2} + x y + y^{2}) = x^{3} - y^{3}$ .

Fig. 4.14Grid mein diagonal pairs cancel ho jaate hain, sirf

x^{3}

aur

- y^{3}

bachte hain.

Step 1 mein ek khaali grid hai: rows ke upar x aur minus y, columns ke upar x squared, xy aur y squared. Step 2 chhah cells mein products bharta hai: x cubed, x squared y, x y squared, minus x squared y, minus x y squared aur minus y cubed. Step 3 do diagonal pairs ko jodta hai, x squared y ko minus x squared y se aur x y squared ko minus x y squared se, unhe halka kar deta hai kyunki har pair ka sum zero hai, aur corner cells x cubed aur minus y cubed ko highlight karta hai. Step 4 result likhta hai: (x minus y)(x squared plus xy plus y squared) barabar x cubed minus y cubed.

TheoremDifference of two cubes

Saare numbers $x$ aur $y$ ke liye, $x^{3} - y^{3} = (x - y) (x^{2} + x y + y^{2})$ .

Ab signs badal do: tumhe kya lagta hai, $(x + y) (x^{2} - x y + y^{2})$ simplify hokar kya banega? Ek guess karo, use $x = 2$ aur $y = 1$ se test karo, aur phir pakka karne ke liye expand karo.

TheoremSum of two cubes

Saare numbers $x$ aur $y$ ke liye, $x^{3} + y^{3} = (x + y) (x^{2} - x y + y^{2})$ .

Ek aur identity teen cubes ko unke product se jodti hai. Yeh lambi dikhti hai, lekin proof bas dhyaan se hisaab rakhna hai.

TheoremThree cubes

Saare numbers $x$ , $y$ aur $z$ ke liye, $x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x)$ .

Example 15NCERT Example 15 · p. 85

Sums aur products se cubes

Teen numbers ka sum 9 hai. Unke pairs ke products ka sum 26 hai, aur unka product 24 hai. Numbers nikale bina unke cubes ka sum nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

4.8 Rational Expressions ko Simplify Karna

$\frac{18}{24}$ jaise fraction ko common factor pehchan kar simplify karte hain: $\frac{18}{24} = \frac{3 \times 6}{4 \times 6} = \frac{3}{4}$ . rational expression (परिमेय व्यंजक) aisa fraction hai jiska numerator aur denominator polynomials hon, jaise $\frac{x^{2} + 5 x + 6}{x^{2} + 4 x + 3}$ . Ise bhi isi tarah simplify karte hain, aur hamari identities hi common factors dikhati hain.

Definition

Rational expression

Aisa fraction jiska numerator aur denominator polynomials hon; yeh sirf wahan defined hai jahan denominator zero nahi hai.

Example $\frac{x + 2}{x - 5}$ , jab $x \neq 5$ .

Fig. 4.15 aise ek fraction ki picture dikhata hai. Dono tile rectangles ki height same hai, $x + 3$ , isliye unke areas ka ratio wahi hai jo unki widths ka:

3/3 $\frac{(x + 2) (x + 3)}{(x + 1) (x + 3)} = \frac{x + 2}{x + 1}$ .

Fig. 4.15Height same hai, isliye areas ka ratio widths ka ratio hai.

Step 1 mein algebra tiles se bane do rectangles hain, jin par x squared plus 5x plus 6 aur x squared plus 4x plus 3 likha hai. Step 2 unki widths, x plus 2 aur x plus 1, mark karta hai aur unki barabar left sides highlight karta hai, har ek ki height x plus 3. Step 3 dono areas ka fraction likhta hai, upar aur neeche factorise karta hai, common factor x plus 3 ko rang se alag dikhata hai, aur use cancel karke (x + 2) over (x + 1) chhodta hai.

\frac{x^{2} + 5 x + 6}{x^{2} + 4 x + 3} = \frac{(x + 2) (x + 3)}{(x + 1) (x + 3)} = \frac{x + 2}{x + 1}

Ek shart hai. Hum upar aur neeche ko $x + 3$ se tabhi divide kar sakte hain jab $x + 3 \neq 0$ ho, kyunki zero se divide karne ka koi matlab nahi. Original fraction ka denominator bhi non-zero hona chahiye, isliye $x \neq - 1$ aur $x \neq - 3$ . In do values ko chhod kar dono sides barabar hain.

Example 16NCERT Example 16 · p. 86

Do factorisations se simplify karna

$\frac{4 x^{2} - 9}{2 x^{2} + 5 x + 3}$ simplify karo, yeh maan kar ki denominator zero nahi hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Practice · 1 question

Practice 4.5: answers ke saath

1
Short answerMediumNCERT Exercise Set 4.5, Q1
Simplify karo, yeh maan kar ki koi denominator zero nahi hai.
1. $\frac{2 m^{2} + 2 m n - 12 n^{2}}{m^{2} + 5 m n + 6 n^{2}}$
  Solution dikhao
  Answer $\frac{2 (m - 2 n)}{m + 2 n}$
  Numerator: $2 (m^{2} + m n - 6 n^{2}) = 2 (m + 3 n) (m - 2 n)$ . Denominator: $(m + 2 n) (m + 3 n)$ . $m + 3 n$ cancel karo to $\frac{2 (m - 2 n)}{m + 2 n}$ milta hai.
2. $\frac{8 a^{3} - 12 a^{2} b + 6 a b^{2} - b^{3}}{4 a^{2} - b^{2}}$
  Solution dikhao
  Answer $\frac{(2 a - b)^{2}}{2 a + b}$
  Numerator $(2 a)^{3} - 3 (2 a)^{2} b + 3 (2 a) b^{2} - b^{3} = (2 a - b)^{3}$ hai, aur denominator $(2 a - b) (2 a + b)$ . $2 a - b$ ek baar cancel karo to $\frac{(2 a - b)^{2}}{2 a + b}$ bachta hai.
3. $\frac{a^{3} + 8 b^{3} + 27 c^{3} - 18 a b c}{a + 2 b + 3 c}$
  Solution dikhao
  Answer $a^{2} + 4 b^{2} + 9 c^{2} - 2 a b - 6 b c - 3 c a$
  Three-cubes identity ko $x = a$ , $y = 2 b$ aur $z = 3 c$ ke saath use karo. Tab $3 x y z = 18 a b c$ , isliye numerator $(a + 2 b + 3 c) (a^{2} + 4 b^{2} + 9 c^{2} - 2 a b - 6 b c - 3 c a)$ hai. $a + 2 b + 3 c$ se divide karne par doosra bracket bachta hai.
4. $\frac{16 a^{2} - 24 a b + 9 b^{2}}{16 a^{2} - 9 b^{2}}$
  Solution dikhao
  Answer $\frac{4 a - 3 b}{4 a + 3 b}$
  $\frac{(4 a - 3 b)^{2}}{(4 a - 3 b) (4 a + 3 b)} = \frac{4 a - 3 b}{4 a + 3 b}$ .
5. $\frac{(y^{2} + 2 y - 15) (y^{2} - 4)}{(y^{2} + 3 y - 10) (y^{2} - 5 y + 6)}$
  Solution dikhao
  Answer $\frac{y + 2}{y - 2}$
  Numerator: $(y + 5) (y - 3) (y - 2) (y + 2)$ . Denominator: $(y + 5) (y - 2) (y - 2) (y - 3)$ . $y + 5$ , $y - 3$ aur ek $y - 2$ cancel karo to $\frac{y + 2}{y - 2}$ bachta hai.
6. $\frac{k^{4} - 81}{k^{3} + 27}$
  Solution dikhao
  Answer $\frac{(k - 3) (k^{2} + 9)}{k^{2} - 3 k + 9}$
  $k^{4} - 81 = (k^{2} - 9) (k^{2} + 9) = (k - 3) (k + 3) (k^{2} + 9)$ aur, sum of two cubes ki tarah, $k^{3} + 27 = (k + 3) (k^{2} - 3 k + 9)$ . $k + 3$ cancel karo.

Factorising practical puzzles bhi suljha deta hai, jaisa agle do examples dikhate hain.

Example 17NCERT Example 17 · p. 87

Floor par carpet bichhana

Ek carpet shop teen size ke mats bechti hai: side $x$ metre ke bade squares, $x$ metre by 1 metre ke runners, aur side 1 metre ke chhote squares. (a) Meena ka floor $(x + 2)$ m by $(x + 6)$ m hai. Kaunse mats bina kaate use poora dhak denge? (b) Uske padosi ke paas 1 bada square, 10 runners aur 21 chhote squares hain. Kya yeh kisi rectangular floor ko poora dhak sakte hain? (c) 1 bada square, 6 runners aur 10 chhote squares ke baare mein kya?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 18NCERT Example 18 · p. 88

Do consecutive even numbers

Do consecutive even numbers ka product 168 hai. Numbers nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Isse hamara toolkit poora hua. Har identity ek picture ya pattern se shuru hui aur ek aise proof par khatam hui jo har number ke liye sahi hai. Saath milkar yeh hamein jaldi expand karne, bharose se factorise karne aur polynomials ke fractions simplify karne mein madad karti hain.

Recap

Summary aur key points

Identity apne letters ki har value ke liye sahi hoti hai; equation sirf kuch khaas values ke liye sahi hoti hai, jinhe hum solve karke dhoondhte hain (Section 4.2).
$(a + b)^{2} = a^{2} + 2 a b + b^{2}$ aur $(a - b)^{2} = a^{2} - 2 a b + b^{2}$ : ek square do squares aur do barabar strips mein bat jaata hai (Section 4.2, Section 4.3).
Ulta padhne par identities factorise karti hain. Trinomial tab perfect square hai jab uske end terms squares hon aur middle term unke roots ke product ka do guna ho (Section 4.3).
$(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a$ (Section 4.4).
$a^{2} - b^{2} = (a + b) (a - b)$ , aur iska rearranged form $a^{2} = (a + b) (a - b) + b^{2}$ Śrīdharācārya ke quick squares deta hai (Section 4.4).
Identities arithmetic ko tez banati hain: number ko ek round hisse aur ek chhote hisse mein todo (Section 4.2).
Algebra tiles ek rectangle banakar $x^{2} + b x + c$ ko factorise karti hain; uski side lengths hi factors hain (Section 4.5).
$(x + a) (x + b) = x^{2} + (a + b) x + a b$ aur $(p x + a) (q x + b) = p q x^{2} + (p b + q a) x + a b$ (Section 4.5).
Tiles ke bina middle term split karo: aise do numbers dhoondho jinka product pehla coefficient times last ho aur sum middle coefficient ho (Section 4.6).
$(a \pm b)^{3} = a^{3} \pm 3 a^{2} b + 3 a b^{2} \pm b^{3}$ aur $x^{3} \pm y^{3} = (x \pm y) (x^{2} \mp x y + y^{2})$ (Section 4.7).
$x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x)$ , isliye jab bhi $x + y + z = 0$ ho, $x^{3} + y^{3} + z^{3} = 3 x y z$ (Section 4.7).
Rational expression ko factorise karke aur common factors cancel karke simplify karte hain, terms kabhi nahi, aur sirf woh factors jo zero nahi hain (Section 4.8).

Flash cards · 11

Identity flashcards: expand karna hai ya factorise?

1 / 11

Chapter ke end ki exercises

Practice · 17 questions

Practice questions answers ke saath (NCERT “Chapter ke end ki exercises”)

1
Short answerMediumNCERT End-of-chapter Exercises, Q1
Har product ko ek identity ki madad se expand karo.
1. $(- 2 y + 5)^{2}$
  Solution dikhao
  Answer $4 y^{2} - 20 y + 25$
  $(- 2 y)^{2} + 2 (- 2 y) (5) + 5^{2} = 4 y^{2} - 20 y + 25$ .
2. $(3 a + 8) (3 a - 8)$
  Solution dikhao
  Answer $9 a^{2} - 64$
  Difference of squares: $(3 a)^{2} - 8^{2} = 9 a^{2} - 64$ .
3. $(\frac{k^{2}}{2} + 3) (\frac{k^{2}}{2} - 3)$
  Solution dikhao
  Answer $\frac{k^{4}}{4} - 9$
  Difference of squares: ${(\frac{k^{2}}{2})}^{2} - 3^{2} = \frac{k^{4}}{4} - 9$ .
4. $(x + 3 y) (x^{2} - 3 x y + 9 y^{2})$
  Solution dikhao
  Answer $x^{3} + 27 y^{3}$
  $y$ ki jagah $3 y$ ke saath sum of cubes: $x^{3} + (3 y)^{3} = x^{3} + 27 y^{3}$ .
5. ${(\frac{1}{3 t} - 3 t)}^{2}$
  Solution dikhao
  Answer $\frac{1}{9 t^{2}} - 2 + 9 t^{2}$
  $\frac{1}{9 t^{2}} - 2 \cdot \frac{1}{3 t} \cdot 3 t + 9 t^{2} = \frac{1}{9 t^{2}} - 2 + 9 t^{2}$ .
6. $(- 2 p + q - 3 r)^{2}$
  Solution dikhao
  Answer $4 p^{2} + q^{2} + 9 r^{2} - 4 p q - 6 q r + 12 p r$
  Squares $4 p^{2}$ , $q^{2}$ , $9 r^{2}$ ; pairs $2 (- 2 p) (q) = - 4 p q$ , $2 (q) (- 3 r) = - 6 q r$ , $2 (- 3 r) (- 2 p) = 12 p r$ .
7. ${(a - \frac{b}{2})}^{3}$
  Solution dikhao
  Answer $a^{3} - \frac{3}{2} a^{2} b + \frac{3}{4} a b^{2} - \frac{1}{8} b^{3}$
  $a^{3} - 3 a^{2} \cdot \frac{b}{2} + 3 a \cdot \frac{b^{2}}{4} - \frac{b^{3}}{8} = a^{3} - \frac{3}{2} a^{2} b + \frac{3}{4} a b^{2} - \frac{1}{8} b^{3}$ .
8. ${(\frac{3}{2} u + \frac{2}{3} v)}^{3}$
  Solution dikhao
  Answer $\frac{27}{8} u^{3} + \frac{9}{2} u^{2} v + 2 u v^{2} + \frac{8}{27} v^{3}$
  ${(\frac{3}{2} u)}^{3} + 3 {(\frac{3}{2} u)}^{2} (\frac{2}{3} v) + 3 (\frac{3}{2} u) {(\frac{2}{3} v)}^{2} + {(\frac{2}{3} v)}^{3} = \frac{27}{8} u^{3} + \frac{9}{2} u^{2} v + 2 u v^{2} + \frac{8}{27} v^{3}$ .
2
NumericalMediumNCERT End-of-chapter Exercises, Q2
Sahi identity use karke har value nikalo.
1. $28 \times 32$
  Answer dekho
  Answer 896
  $(30 - 2) (30 + 2) = 900 - 4 = 896$ .
2. $203 \times 197$
  Answer dekho
  Answer 39991
  $(200 + 3) (200 - 3) = 40000 - 9 = 39991$ .
3. $45 \times 35$
  Answer dekho
  Answer 1575
  $(40 + 5) (40 - 5) = 1600 - 25 = 1575$ .
4. $103^{3}$
  Answer dekho
  Answer 1092727
  $(100 + 3)^{3} = 1000000 + 90000 + 2700 + 27 = 1092727$ .
5. $99^{3}$
  Answer dekho
  Answer 970299
  $(100 - 1)^{3} = 1000000 - 30000 + 300 - 1 = 970299$ .
6. $201^{3}$
  Answer dekho
  Answer 8120601
  $(200 + 1)^{3} = 8000000 + 120000 + 600 + 1 = 8120601$ .
7. $(- 101)^{3}$
  Answer dekho
  Answer -1030301
  $(- 101)^{3} = - (100 + 1)^{3} = - (1000000 + 30000 + 300 + 1) = - 1030301$ .
8. $(- 198)^{3}$
  Answer dekho
  Answer -7762392
  $(- 198)^{3} = - (200 - 2)^{3} = - (8000000 - 240000 + 2400 - 8) = - 7762392$ .
3
Short answerMediumNCERT End-of-chapter Exercises, Q3
Yeh do aur teen terms wale expressions factorise karo.
1. $4 p^{2} + 4 + \frac{1}{p^{2}}$
  Solution dikhao
  Answer ${(2 p + \frac{1}{p})}^{2}$
  $2 \times 2 p \times \frac{1}{p} = 4$ , isliye yeh ${(2 p + \frac{1}{p})}^{2}$ hai.
2. $25 a^{2} - \frac{1}{36 b^{2}}$
  Solution dikhao
  Answer $(5 a - \frac{1}{6 b}) (5 a + \frac{1}{6 b})$
  Difference of squares: $(5 a)^{2} - {(\frac{1}{6 b})}^{2}$ .
3. $125 k^{3} - \frac{1}{8 k^{3}}$
  Solution dikhao
  Answer $(5 k - \frac{1}{2 k}) (25 k^{2} + \frac{5}{2} + \frac{1}{4 k^{2}})$
  $x = 5 k$ aur $y = \frac{1}{2 k}$ ke saath difference of cubes: $x^{2} + x y + y^{2} = 25 k^{2} + \frac{5}{2} + \frac{1}{4 k^{2}}$ .
4. $y^{2} + \frac{3 y}{10} + \frac{1}{50}$
  Solution dikhao
  Answer $(y + \frac{1}{5}) (y + \frac{1}{10})$
  Product $\frac{1}{50}$ aur sum $\frac{3}{10}$ wale do numbers $\frac{1}{5}$ aur $\frac{1}{10}$ hain.
5. $8 m^{3} + \frac{n^{3}}{27}$
  Solution dikhao
  Answer $(2 m + \frac{n}{3}) (4 m^{2} - \frac{2 m n}{3} + \frac{n^{2}}{9})$
  Sum of two cubes, jismein $x$ aur $y$ ki jagah $2 m$ aur $\frac{n}{3}$ hain.
6. $16 h^{2} - 24 h + 9$
  Solution dikhao
  Answer $(4 h - 3)^{2}$
  $(4 h)^{2} - 2 (4 h) (3) + 3^{2}$ .
4
Short answerHardNCERT End-of-chapter Exercises, Q3
Yeh lambe expressions factorise karo.
1. $6 w - 8 + \frac{w^{3}}{8} - \frac{3 w^{2}}{2}$
  Solution dikhao
  Answer ${(\frac{w}{2} - 2)}^{3}$
  $w$ ki powers ke hisaab se lagao: $\frac{w^{3}}{8} - \frac{3 w^{2}}{2} + 6 w - 8$ . $A = \frac{w}{2}$ aur $B = 2$ ke saath: $A^{3} = \frac{w^{3}}{8}$ , $3 A^{2} B = \frac{3 w^{2}}{2}$ , $3 A B^{2} = 6 w$ aur $B^{3} = 8$ . Signs baari-baari badalte hain, isliye yeh $(A - B)^{3}$ hai.
2. $a^{3} + 8 b^{3} + c^{3} - 6 a b c$
  Solution dikhao
  Answer $(a + 2 b + c) (a^{2} + 4 b^{2} + c^{2} - 2 a b - 2 b c - c a)$
  $x = a$ , $y = 2 b$ , $z = c$ ke saath three cubes: $3 x y z = 6 a b c$ .
3. $2 x^{3} - 16 y^{3} - 54 z^{3} - 36 x y z$
  Solution dikhao
  Answer $2 (x - 2 y - 3 z) (x^{2} + 4 y^{2} + 9 z^{2} + 2 x y - 6 y z + 3 z x)$
  2 bahar nikalo: $x^{3} + (- 2 y)^{3} + (- 3 z)^{3} - 3 x (- 2 y) (- 3 z)$ . $- 2 y$ aur $- 3 z$ ke saath three-cubes identity jawab deti hai.
4. $9 x^{2} + 4 y^{2} + 16 z^{2} + 12 x y + 16 y z + 24 z x$
  Solution dikhao
  Answer $(3 x + 2 y + 4 z)^{2}$
  Saare cross terms positive hain: $2 (3 x) (2 y) = 12 x y$ , $2 (2 y) (4 z) = 16 y z$ , $2 (4 z) (3 x) = 24 z x$ .
5. $64 p^{3} - 48 p^{2} + 12 p - 1$
  Solution dikhao
  Answer $(4 p - 1)^{3}$
  $(4 p)^{3} - 3 (4 p)^{2} (1) + 3 (4 p) (1)^{2} - 1^{3}$ .
5
Short answerMediumNCERT End-of-chapter Exercises, Q4
Simplify karo, yeh maan kar ki denominators zero nahi hain.
1. $\frac{9 x^{2} - 1}{3 x^{2} + 2 x - 1}$
  Solution dikhao
  Answer $\frac{3 x + 1}{x + 1}$
  Numerator: $(3 x - 1) (3 x + 1)$ . Denominator: product $3 \times (- 1) = - 3$ aur sum 2 se numbers 3 aur $- 1$ milte hain, isliye $3 x^{2} + 3 x - x - 1 = 3 x (x + 1) - (x + 1) = (3 x - 1) (x + 1)$ . $3 x - 1$ cancel karo to $\frac{3 x + 1}{x + 1}$ bachta hai.
2. $\frac{2 x^{3} - 54 y^{3}}{x^{2} - 6 x y + 9 y^{2}}$
  Solution dikhao
  Answer $\frac{2 (x^{2} + 3 x y + 9 y^{2})}{x - 3 y}$
  $\frac{2 (x - 3 y) (x^{2} + 3 x y + 9 y^{2})}{(x - 3 y)^{2}} = \frac{2 (x^{2} + 3 x y + 9 y^{2})}{x - 3 y}$ .
3. $\frac{m^{3} + 64 n^{3}}{m^{2} - 16 n^{2}}$
  Solution dikhao
  Answer $\frac{m^{2} - 4 m n + 16 n^{2}}{m - 4 n}$
  $\frac{(m + 4 n) (m^{2} - 4 m n + 16 n^{2})}{(m + 4 n) (m - 4 n)} = \frac{m^{2} - 4 m n + 16 n^{2}}{m - 4 n}$ .
6
Short answerMediumNCERT End-of-chapter Exercises, Q5
Neeche har expression ek rectangle ka area hai, square units mein. Ise factorise karke ek length aur ek breadth suggest karo.
1. $9 p^{2} - 42 p q + 49 q^{2}$
  Solution dikhao
  AnswerDono sides $3 p - 7 q$ (yeh ek square hai).
  $9 p^{2} - 42 p q + 49 q^{2} = (3 p - 7 q)^{2}$ , isliye rectangle side $3 p - 7 q$ , ya $7 q - 3 p$ , jo bhi positive ho, ka square hai.
2. $64 m^{2} - 25 n^{2}$
  Solution dikhao
  Answer $8 m + 5 n$ aur $8 m - 5 n$
  Difference of squares: $(8 m + 5 n) (8 m - 5 n)$ .
7
Short answerMediumNCERT End-of-chapter Exercises, Q6
Neeche har expression ek cuboid ka volume hai, cubic units mein. Ise factorise karke ek length, ek breadth aur ek height suggest karo.
1. $7 m^{2} - 63 n^{2}$
  Solution dikhao
  Answer $7$ , $m - 3 n$ aur $m + 3 n$
  $7 (m^{2} - 9 n^{2}) = 7 (m - 3 n) (m + 3 n)$ .
2. $4 h r^{2} - 20 h r + 24 h$
  Solution dikhao
  Answer $4 h$ , $r - 2$ aur $r - 3$
  $4 h (r^{2} - 5 r + 6) = 4 h (r - 2) (r - 3)$ .
Short answerMediumNCERT End-of-chapter Exercises, Q7
Tin ki ek square sheet ki sides 30 cm hain. Har corner se side $w$ cm ka ek square kaata jaata hai taaki edges ko mod kar ek khula box banaya ja sake. Bachi hui tin ka area $w$ ke terms mein likho, aur use factorise karo.
Solution dikhao
Answer $900 - 4 w^{2} = 4 (15 - w) (15 + w)$ square centimetres
Bachi hui tin $30^{2} - 4 w^{2} = 900 - 4 w^{2} = 4 (225 - w^{2}) = 4 (15 - w) (15 + w)$ square centimetres hai.
Short answerMediumNCERT End-of-chapter Exercises, Q8
Ek number ka do guna, minus uske reciprocal ka teen guna, 5 aata hai. Number kya ho sakta hai?
Solution dikhao
Answer $3$ ya $- \frac{1}{2}$
$2 x - \frac{3}{x} = 5$ se $2 x^{2} - 5 x - 3 = 0$ milta hai. $- 5 x$ ko $- 6$ aur $1$ se split karo (product $2 \times (- 3) = - 6$ , sum $- 5$ ): $2 x^{2} - 6 x + x - 3 = 2 x (x - 3) + (x - 3) = (2 x + 1) (x - 3) = 0$ .
Isliye $x = 3$ ya $x = - \frac{1}{2}$ . Check: $6 - 1 = 5$ , aur $- 1 - (- 6) = 5$ .
Short answerMediumNCERT End-of-chapter Exercises, Q9
Ek rectangular solar panel ka area $3 x^{2} + 11 x + 6$ square decimetres hai, aur uski ek side $x + 3$ decimetres hai. Doosri side nikalo.
Solution dikhao
Answer $3 x + 2$ decimetres
$3 \times 6 = 18$ ; product 18 aur sum 11 wale numbers 2 aur 9 hain. Isliye $3 x^{2} + 2 x + 9 x + 6 = x (3 x + 2) + 3 (3 x + 2) = (x + 3) (3 x + 2)$ . Doosri side $3 x + 2$ decimetres hai.
Short answerHardNCERT End-of-chapter Exercises, Q10
$x + 2$ aur $3 x - 1$ dono $p x^{2} + q x - 4$ ke factors hain. $p$ aur $q$ nikalo.
Solution dikhao
Answer $p = 6$ , $q = 10$
$(x + 2) (3 x - 1) = 3 x^{2} + 5 x - 2$ , isliye kisi number $k$ ke liye $p x^{2} + q x - 4 = k (3 x^{2} + 5 x - 2)$ .
Constant terms se $- 2 k = - 4$ milta hai, isliye $k = 2$ . Tab $p = 3 k = 6$ aur $q = 5 k = 10$ : quadratic $6 x^{2} + 10 x - 4$ hai.
ProofHardNCERT End-of-chapter Exercises, Q11
Agar $a + b + c = 7$ aur $a^{2} + b^{2} + c^{2} = 21$ ho, to prove karo ki $a^{3} + b^{3} + c^{3} - 3 a b c = 49$ .
Solution dikhao
$49 = (a + b + c)^{2} = 21 + 2 (a b + b c + c a)$ , isliye $a b + b c + c a = 14$ .
Three-cubes identity se, $a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - (a b + b c + c a)) = 7 (21 - 14) = 49$ .
ProofHardNCERT End-of-chapter Exercises, Q12
Factorise karke dikhao ki har natural number $n$ ke liye $n^{3} + 3 n^{2} + 2 n$ , 6 se divisible hai.
Solution dikhao
$n^{3} + 3 n^{2} + 2 n = n (n^{2} + 3 n + 2) = n (n + 1) (n + 2)$ , yaani teen consecutive natural numbers ka product.
Kinhi bhi do consecutive numbers mein ek even hota hai, isliye product 2 se divisible hai. Kinhi bhi teen consecutive numbers mein ek 3 ka multiple hota hai, isliye product 3 se divisible hai. 2 aur 3 dono se divisible hone ke kaaran yeh 6 se divisible hai.
14
ProofHardNCERT End-of-chapter Exercises, Q13
Har value nikalo.
1. $m^{3} - n^{3} - 8 - 6 m n$ , jab $m - n = 2$
  Answer dekho
  Answer 0
  $x = m$ , $y = - n$ aur $z = - 2$ lo. Tab $x + y + z = m - n - 2 = 0$ aur $3 x y z = 3 m (- n) (- 2) = 6 m n$ , isliye expression $x^{3} + y^{3} + z^{3} - 3 x y z = 0$ hai.
2. $13^{3} - 8^{3} - 5^{3}$ , koi bhi cube nikale bina
  Answer dekho
  Answer 1560
  $13 + (- 8) + (- 5) = 0$ , isliye teeno cubes ka sum $3 \times 13 \times (- 8) \times (- 5) = 1560$ hai. Check: $2197 - 512 - 125 = 1560$ .
Quick answerEasy
Inmein se kaunsa identity nahi hai?
$(x - 1)^{2} = x^{2} - 2 x + 1$ $(x - 1) (x + 1) = x^{2} - 1$ $(x - 1)^{3} = x^{3} - 1$ $x^{3} - 1 = (x - 1) (x^{2} + x + 1)$
Answer dekho
Answer (C) $(x - 1)^{3} = x^{3} - 1$
$(x - 1)^{3} = x^{3} - 3 x^{2} + 3 x - 1$ ; $x = 2$ par yeh 1 hai, jabki $x^{3} - 1 = 7$ . Baaki teeno is chapter ki identities hain.
16
Quick answerMedium
Har expression ko uske factorised form se match karo.
Left mein koi item tap karo, phir right mein uska match.
Answer dekho
Answer
$x^{2} - 9$ : $(x - 3) (x + 3)$
$x^{2} - 6 x + 9$ : $(x - 3)^{2}$
$x^{3} - 27$ : $(x - 3) (x^{2} + 3 x + 9)$
$x^{2} + 3 x - 18$ : $(x + 6) (x - 3)$
Difference of squares, square of a difference, difference of cubes aur middle term ko split karna (product $- 18$ , sum 3).
17
Quick answerMedium
$2 x^{2} + 7 x + 6$ ko factorise karne ke steps ko order mein lagao.
1. $2 x^{2} + 3 x + 4 x + 6$ ki tarah dobara likho
2. Bahar wale coefficients multiply karo: $2 \times 6 = 12$
3. 3 aur 4 dhoondho: product 12, sum 7
4. Group karo: $x (2 x + 3) + 2 (2 x + 3)$
5. $(2 x + 3) (x + 2)$ likho
Answer dekho
Answer
Bahar wale coefficients multiply karo: $2 \times 6 = 12$
3 aur 4 dhoondho: product 12, sum 7
$2 x^{2} + 3 x + 4 x + 6$ ki tarah dobara likho
Group karo: $x (2 x + 3) + 2 (2 x + 3)$
$(2 x + 3) (x + 2)$ likho
Pehle product, phir pair, phir split, grouping aur aakhir mein common bracket.

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercise Set 4.1, Q1	Practice 4.1 · Q1
NCERT Exercise Set 4.1, Q2	Practice 4.1 · Q2
NCERT Exercise Set 4.2, Q1	Practice 4.2 · Q1
NCERT Exercise Set 4.2, Q2	Practice 4.2 · Q2
NCERT Exercise Set 4.3, Q1	Practice 4.3 · Q1
NCERT Exercise Set 4.3, Q2	Practice 4.3 · Q2
NCERT Exercise Set 4.3, Q3	Practice 4.3 · Q3
NCERT Exercise Set 4.3, Q4	Practice 4.3 · Q4
NCERT Exercise Set 4.4, Q1	Practice 4.4 · Q1
NCERT Exercise Set 4.4, Q2	Practice 4.4 · Q2
NCERT Exercise Set 4.4, Q3	Practice 4.4 · Q3
NCERT Exercise Set 4.5, Q1	Practice 4.5 · Q1
NCERT End-of-chapter Exercises, Q1	Chapter ke end ki exercises · Q1
NCERT End-of-chapter Exercises, Q2	Chapter ke end ki exercises · Q2
NCERT End-of-chapter Exercises, Q3	Chapter ke end ki exercises · Q3
NCERT End-of-chapter Exercises, Q3	Chapter ke end ki exercises · Q4
NCERT End-of-chapter Exercises, Q4	Chapter ke end ki exercises · Q5
NCERT End-of-chapter Exercises, Q5	Chapter ke end ki exercises · Q6
NCERT End-of-chapter Exercises, Q6	Chapter ke end ki exercises · Q7
NCERT End-of-chapter Exercises, Q7	Chapter ke end ki exercises · Q8
NCERT End-of-chapter Exercises, Q8	Chapter ke end ki exercises · Q9
NCERT End-of-chapter Exercises, Q9	Chapter ke end ki exercises · Q10
NCERT End-of-chapter Exercises, Q10	Chapter ke end ki exercises · Q11
NCERT End-of-chapter Exercises, Q11	Chapter ke end ki exercises · Q12
NCERT End-of-chapter Exercises, Q12	Chapter ke end ki exercises · Q13
NCERT End-of-chapter Exercises, Q13	Chapter ke end ki exercises · Q14

Quick check: MCQs

4.1 Introduction3 questions
4.2 Identities ko Visualise Karna4 questions
4.3 Identities ki madad se Algebraic Expressions ka Factorisation4 questions
4.4 Aur Identities4 questions
4.5 Algebra Tiles se Factorisation4 questions
4.6 Algebra Tiles ke Bina Factorisation4 questions
4.7 Nayi Identities Dhoondhna4 questions
4.8 Rational Expressions ko Simplify Karna4 questions

Glossary · 10

Key terms (English + Hindi)

algebra tiles: Areas $x^{2}$ , $x$ aur 1 wale squares aur strips, jinse products aur factors ka model banta hai.Chapter mein pehli baar yahan aaya →

coefficientगुणांकgunank: Term mein letters ko multiply karne wala number; $7 x^{2}$ mein yeh 7 hai.Chapter mein pehli baar yahan aaya →

difference of two squaresदो वर्गों का अंतरdo vargon ka antar: Expression $a^{2} - b^{2}$ , jo $(a + b) (a - b)$ mein factorise hota hai.Chapter mein pehli baar yahan aaya →

equationसमीकरणsamikaran: Aisa statement ki do expressions barabar hain. $3 x - 5 = 16$ jaisa normal equation sirf kuch khaas values ke liye sahi hota hai, jo use solve karke milti hain; identity saari values ke liye sahi hoti hai.Chapter mein pehli baar yahan aaya →

factorगुणनखंडgunankhand: Aisa expression jo doosre ko poora-poora divide kare; $x + 2$ , $x^{2} + 5 x + 6$ ka ek factor hai.Chapter mein pehli baar yahan aaya →
factorisationगुणनखंडनgunankhandan: Kisi expression ko simple expressions ke product ki tarah likhna.Chapter mein pehli baar yahan aaya →

identityसर्वसमिकाsarvasamika: Do expressions ki aisi equality jo uske letters ki har value ke liye sahi ho, jaise $(a + b)^{2} = a^{2} + 2 a b + b^{2}$ .Chapter mein pehli baar yahan aaya →

perfect-square trinomialपूर्ण वर्गpoorn varg: Teen terms wala expression $a^{2} \pm 2 a b + b^{2}$ , jo $(a \pm b)^{2}$ ke barabar hai.Chapter mein pehli baar yahan aaya →

rational expressionपरिमेय व्यंजकparimey vyanjak: Aisa fraction jiska numerator aur denominator polynomials hon, aur denominator non-zero ho.Chapter mein pehli baar yahan aaya →

splitting the middle term: $a x^{2} + b x + c$ ko factorise karna, $b x$ ko aise do terms mein likh kar jinke coefficients ka product $a c$ ho.Chapter mein pehli baar yahan aaya →

4.1 Introduction

4.2 Identities ko Visualise Karna

4.3 Identities ki madad se Algebraic Expressions ka Factorisation

4.4 Aur Identities

4.5 Algebra Tiles se Factorisation

4.6 Algebra Tiles ke Bina Factorisation

4.7 Nayi Identities Dhoondhna

4.8 Rational Expressions ko Simplify Karna

Chapter ke end ki exercises

Apna NCERT exercise question dhoondo

Quick check: MCQs

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