Polynomials: Class 10 Maths · Chapter 2

Mathematics · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

60 min
1 simulation
15 practice questions
15 MCQs

What you will learn in Polynomials

Name a polynomial as linear, quadratic or cubic from its degree and write the general form of each.
Find the value p(k) of a polynomial and decide whether k is a zero of it.
Read the zeroes of a polynomial off the graph of y = p(x): they are the values of x at which the curve meets the x-axis.
Explain why a quadratic polynomial has at most two zeroes, a cubic at most three and a polynomial of degree n at most n.
Find the zeroes of a quadratic polynomial by factorising, and verify that their sum is −b/a and their product is c/a.
Write a quadratic polynomial when the sum and the product of its zeroes are given.
Check the three relations between the zeroes and the coefficients of a cubic polynomial.

Original companion text aligned to NCERT Mathematics Class 10, Chapter 2, pp. 10–23. Not affiliated with or endorsed by NCERT.

2.1 Introduction

Suppose a ball leaves the ground going straight up at 20 m/s. Ignore air resistance and take $g \approx 10 m / s^{2} .$ Then its height after $t$ seconds is about $h (t) = 20 t - 5 t^{2}$ metres. A few values: $h (1) = 15,$ $h (2) = 20$ and $h (3) = 15.$ When is the ball back on the ground? We want the times at which the height is zero. Try $t = 4 :$ $h (4) = 80 - 80 = 0.$ So the ball lands after 4 seconds (Fig. 2.1).

3/3 $h (t) = 0$ at $t = 0$ (launch) and at $t = 4$ (landing).

Fig. 2.1The inputs that make the height zero,

t = 0

and

t = 4,

are the zeroes of

20 t - 5 t^{2} .

A graph of height in metres against time in seconds, with a small table of values on the right. In the first step only the rising half of the curve is drawn, from height 0 at time 0 up to the top point, 20 metres at 2 seconds. In the second step the falling half appears, coming back down to height 0 at 4 seconds. In the third step the two points on the time axis, at 0 and 4 seconds, are ringed and the table rows where the height is 0 are highlighted.

The numbers $0$ and $4$ are special inputs for $20 t - 5 t^{2} :$ they make its value zero. In this chapter you will find such inputs, see them on a graph, and link them to the numbers that appear in the polynomial. First, a quick recap from Class 9.

The degree of a polynomial in one variable is the highest power of the variable in it. Polynomials of degree 1, 2 and 3 have their own names: linear, quadratic and cubic (Table 2.1). In each general form, the letters $a,$ $b,$ $c,$ $d$ are real numbers called coefficients. The leading coefficient $a$ must not be zero, or the degree would drop.

Table 2.1Polynomials named by their degree (

a \neq 0

in each general form)

Degree and name	General form	Examples
1: linear	$a x + b$	$7 - 2 x,$ $\sqrt{3} y + 1$
2: quadratic	$a x^{2} + b x + c$	$x^{2} - 9,$ $5 u - u^{2}$
3: cubic	$a x^{3} + b x^{2} + c x + d$	$x^{3},$ $4 - 2 p^{3} + p$

The word quadratic comes from the Latin quadratus, 'made square': $x^{2}$ is the area of a square of side $x .$ Expressions such as $\frac{1}{x} + 2$ and $\sqrt{x} - 5$ are not polynomials. In them, $x$ sits in a denominator or under a root sign, which means a power such as $- 1$ or $\frac{1}{2}$ that is not a whole number.

Value and zero of a polynomial

Take $p (x) = x^{2} + x - 6.$ Put a number $k$ in place of $x$ and work out the result: that is the value of the polynomial at $k,$ written $p (k) .$ For example, $p (1) = 1 + 1 - 6 = - 4$ and $p (0) = - 6.$ Notice that $p (0)$ is always the constant term. Now try $x = 2$ and $x = - 3 :$ $p (2) = 4 + 2 - 6 = 0$ and $p (- 3) = 9 - 3 - 6 = 0.$

Definition

Zero of a polynomial

A zero of $p (x)$ is any real number $k$ that makes the value of the polynomial zero, that is, $p (k) = 0.$

Example $2$ and $- 3$ are zeroes of $x^{2} + x - 6,$ and $4$ is a zero of $20 t - 5 t^{2} .$

Finding the zero of a linear polynomial is the same as solving a linear equation, which you did in Class 9. For $5 x + 8,$ a zero $k$ must satisfy $5 k + 8 = 0,$ so $k = - \frac{8}{5} .$ In general, if $a k + b = 0$ with $a \neq 0,$ then:

Zero of a linear polynomial

k = - \frac{b}{a}

(2.1)

where

$b$: the constant term
$a$: the coefficient of $x$

Notice that $- \frac{b}{a}$ uses nothing but the two coefficients. Could the zeroes of a quadratic, which may have two of them, also be predicted from $a,$ $b$ and $c$ ? We return to this in Section 2.3. First, let us see what zeroes look like on a graph.

Quick check

1 of 4

Which of these is a quadratic polynomial?

x^{3} - x^{2}

4 - 3 x^{2} + x

x^{2} + \frac{1}{x}

2 x + 7

Show answer

Answer (B) $4 - 3 x^{2} + x$

4 - 3 x^{2} + x

has degree 2.

x^{3} - x^{2}

has degree 3, and

x^{2} + \frac{1}{x}

is not a polynomial at all, because

\frac{1}{x}

x

to the power

- 1.

2.2 Geometrical Meaning of the Zeroes of a Polynomial

What does a zero look like on a graph? In Fig. 2.1 the zeroes told us when the ball was on the ground. A graph shows the same thing for any polynomial. On the graph of $y = p (x),$ the value $p (k)$ is the height of the curve above or below the $x$ -axis at $x = k .$ A zero is an input where that height is exactly 0, so at a zero the curve is on the $x$ -axis. Let us test this idea on linear, quadratic and cubic polynomials.

Linear polynomials

In Class 9 you drew the graph of $y = a x + b,$ a straight line. Take $y = 4 - 2 x .$ Some points on it are $(0, 4),$ $(1, 2),$ $(2, 0)$ and $(3, - 2) .$ The line slopes down and crosses the $x$ -axis at $(2, 0)$ (Fig. 2.2). And $2$ is exactly the zero of $4 - 2 x,$ since $4 - 2 (2) = 0.$

3/3On the $x$ -axis $y = 0,$ so $4 - 2 x = 0$ and $x = 2.$

Fig. 2.2The zero of

4 - 2 x

is where its line crosses the

x

-axis.NCERT Fig. 2.1 · p. 12

A square grid with x from minus 1 to 5 and y from minus 3 to 5. In the first step the falling line y equals 4 minus 2x is drawn through four marked points: (0, 4), (1, 2), (2, 0) and (3, minus 2). In the second step the point (2, 0) where the line crosses the x-axis is ringed and labelled, with the note that the line meets the x-axis once. In the third step the working appears on the right: on the x-axis y is 0, so 4 minus 2x equals 0 and x equals 2.

The same reasoning works for any line $y = a x + b$ with $a \neq 0.$ Points on the $x$ -axis have $y = 0,$ so the crossing point must satisfy $a x + b = 0,$ which gives $x = - \frac{b}{a} .$ A line that is not flat cuts the $x$ -axis just once, so every linear polynomial has exactly one zero, at the crossing point $(- \frac{b}{a}, 0) .$

Quadratic polynomials

Now take the quadratic $p (x) = x^{2} + 2 x - 3.$ Table 2.2 lists the values of $y = x^{2} + 2 x - 3$ for the whole numbers from $- 4$ to $2.$

Table 2.2Values of

y = x^{2} + 2 x - 3

NCERT Table 2.1 · p. 13

$x$	$- 4$	$- 3$	$- 2$	$- 1$	$0$	$1$	$2$
$y$	$5$	$0$	$- 3$	$- 4$	$- 3$	$0$	$5$

Plot these seven points and join them with a smooth curve: you get the U-shape in Fig. 2.3. The table shows $y = 0$ at $x = - 3$ and at $x = 1,$ so these are the zeroes. On the graph they are exactly where the curve meets the $x$ -axis, at $(- 3, 0)$ and $(1, 0) .$

3/3The curve meets the $x$ -axis at $(- 3, 0)$ and $(1, 0) .$

Fig. 2.3The zeroes of

x^{2} + 2 x - 3

are

- 3

and

1 :

the curve meets the

x

-axis at exactly those values of

x .

NCERT Fig. 2.2 · p. 13

A grid with x from minus 5 to 3 and y from minus 5 to 6, and notes on the right. In the first step only the seven points from the table are plotted, from (minus 4, 5) down to (minus 1, minus 4) and back up to (2, 5). In the second step a smooth U-shaped curve joins them. In the third step the two points where the curve meets the x-axis, (minus 3, 0) and (1, 0), are ringed and labelled, and the note says the zeroes are minus 3 and 1.

SimulationPlot the table yourself

An empty square grid from minus 6 to 6 on both axes. Each tap adds a labelled point that snaps to whole numbers, and a list beside the grid gives each point's coordinates and says where it lies, for example on the x-axis or in a quadrant. Points can be dragged to correct them.

Fig. 2.4Tap the grid to plot the points of

y = x^{2} + 2 x - 3

and watch the U-shape appear.

Try this

Plot $(- 4, 5),$ $(- 3, 0),$ $(- 2, - 3),$ $(- 1, - 4),$ $(0, - 3),$ $(1, 0)$ and $(2, 5) .$ Which two points does the list say are on the $x$ -axis?
Which plotted point is the lowest? About which vertical line is the U-shape symmetric, and how is that line related to the two zeroes?

The graph of every quadratic polynomial is a curve of this kind, called a parabola (परवलय). Which way it opens depends on the sign of $a$ in $a x^{2} + b x + c$ (Fig. 2.5). When $a > 0,$ the parabola opens upward like a cup. When $a < 0,$ it opens downward like an upturned bowl. For the ball, $a = - 5,$ which is why its height graph in Fig. 2.1 opens downward.

Definition

Parabola

The curve that is the graph of $y = a x^{2} + b x + c$ with $a \neq 0.$ It opens upward when $a > 0$ and downward when $a < 0.$

ExampleThe graph of $y = x^{2} + 2 x - 3$ opens upward, since $a = 1.$

2/2 $y = 6 + x - x^{2}$ has $a = - 1 < 0,$ so it opens downward.

Fig. 2.5The sign of

a

decides which way a parabola opens.

A grid with x from minus 4 to 4 and y from minus 5 to 5. In the first step the solid parabola y equals x squared minus 4 is drawn; it opens upward and the note reads opens up when a is greater than 0. In the second step a dashed parabola y equals 6 plus x minus x squared appears; it opens downward, the note reads opens down when a is less than 0, and the first parabola fades.

Two zeroes, one zero or none

A parabola and the $x$ -axis can meet in only three ways. Picture the parabola $y = x^{2} - 2 x - 3$ sliding upward, which is what happens when you increase its constant term (Fig. 2.6).

Case (i): two crossings. The curve passes through the $x$ -axis at two separate places; call them $A$ and $B .$ Each gives a zero, so there are two different zeroes. For $x^{2} - 2 x - 3$ they are $- 1$ and $3.$
Case (ii): a single touch. The lowest point of the curve rests on the $x$ -axis. Imagine $A$ and $B$ sliding towards each other until they merge. The quadratic then has two equal zeroes, and on the graph we see one zero. For $(x - 1)^{2} = x^{2} - 2 x + 1$ that zero is $1.$
Case (iii): no point. The whole parabola lies above the $x$ -axis (below it, when $a < 0$ ). There is no zero. For example, $x^{2} - 2 x + 3 = (x - 1)^{2} + 2$ is never less than 2.

1/3 $y = x^{2} - 2 x - 3$ cuts the axis at $A (- 1, 0)$ and $B (3, 0) .$

Fig. 2.6A quadratic polynomial has two zeroes, two equal zeroes, or no zero.NCERT Figs. 2.3–2.5 · p. 14

A grid with x from minus 3 to 5 and y from minus 5 to 6, with case notes on the right. In the first step the parabola y equals x squared minus 2x minus 3 cuts the x-axis at A (minus 1, 0) and B (3, 0), both ringed: case (i), two zeroes. In the second step the same curve has slid up 4 units and only touches the x-axis at (1, 0), marked A equals B: case (ii), two equal zeroes. In the third step it has slid up 2 more units and lies wholly above the x-axis: case (iii), no zero.

A downward parabola gives the same three cases, upside down. Either way, a parabola can never meet the $x$ -axis at three points. So a quadratic polynomial has at most two zeroes.

In the Zero explorer (Fig. 2.7) you set $a,$ $b$ and $c,$ and the graph and its zeroes update at once. Try to make each of the three cases, and watch what $a$ does to the shape.

InteractiveZero explorer

An interactive graph on a grid from minus 6 to 6. In the Quadratic tab, sliders and plus and minus buttons set a, b and c in y equals a x squared plus b x plus c, in steps of 0.5. The parabola is redrawn at once, each point where it meets the x-axis is ringed and labelled, and a status box says how many zeroes there are, whether the curve only touches the axis and which way it opens. Example buttons load two zeroes, a touching parabola, no zero and a downward parabola. A Cubic tab does the same for a x cubed plus b x squared plus c x plus d. A checkbox opens a table comparing the sum and product of the zeroes with minus b over a and c over a, and for a cubic the three cubic relations.

Fig. 2.7Change the coefficients and watch the zeroes move. The Cubic tab and the coefficient check are used again in Section 2.3.

Cubic polynomials

What should we expect from a cubic? Take $p (x) = x^{3} - 3 x^{2} - x + 3.$ Its values for the whole numbers from $- 2$ to $4$ are in Table 2.3.

Table 2.3Values of

y = x^{3} - 3 x^{2} - x + 3

NCERT Table 2.2 · p. 15

$x$	$- 2$	$- 1$	$0$	$1$	$2$	$3$	$4$
$y$	$- 15$	$0$	$3$	$0$	$- 3$	$0$	$15$

The value is 0 at $x = - 1,$ $1$ and $3.$ Grouping the terms confirms it: $p (x) = x^{2} (x - 3) - (x - 3),$ which is $(x - 3) (x^{2} - 1),$ and that factorises as $(x - 3) (x - 1) (x + 1) .$

The graph in Fig. 2.8 climbs, turns down, turns up again, and crosses the $x$ -axis at exactly these three points.

Not every cubic meets the axis three times. In $x^{3} - 2 x^{2} = x^{2} (x - 2),$ the factor $x^{2}$ makes $0$ a repeated zero. The graph only touches the axis at the origin and then crosses it at $2,$ so the zeroes are $0$ and $2.$ In $x^{3} + x = x (x^{2} + 1),$ the factor $x^{2} + 1$ is never zero (it is at least 1). So $0$ is the only zero, and the graph crosses just once.

1/3 $x^{3} - 3 x^{2} - x + 3$ crosses at $- 1,$ $1$ and $3.$

Fig. 2.8A cubic polynomial has at least one and at most three zeroes.NCERT Figs. 2.6–2.8 · p. 16

A grid with x from minus 2 to 4 and y from minus 6 to 6, with notes on the right. In the first step the solid curve y equals x cubed minus 3x squared minus x plus 3 crosses the x-axis at minus 1, 1 and 3, all three ringed. In the second step it fades and the dashed curve y equals x cubed minus 2x squared appears; it touches the axis at the origin and crosses at 2, both ringed. In the third step the dotted curve y equals x cubed plus x appears, crossing only at the origin.

Notice two things. A cubic's graph starts far below the axis and ends far above it (or the other way round when $a < 0$ ), so it must cross the axis at least once. And it turns at most twice, so it cannot meet the axis more than three times.

PropertyZeroes and degree

A polynomial of degree $n$ cannot have more than $n$ zeroes: the graph of $y = p (x)$ meets the $x$ -axis $n$ times or fewer.

Example 1NCERT Example 1 · p. 17

Count the zeroes from a graph

Fig. 2.9 shows four curves, each drawn from some polynomial as $y = p (x) .$ Count the zeroes of $p (x)$ for each one.

Fig. 2.9Graphs for Example 1.NCERT Fig. 2.9 · p. 17

Four small graphs in a 2 by 2 grid, each with x and y axes. (a) A cubic curve rises from the bottom left, crosses the x-axis once on the left, then makes a hump and a dip that both stay above the axis. (b) A downward parabola whose top just touches the x-axis. (c) A curve that crosses the x-axis at two points and then just touches it at a third point on the right. (d) An upward parabola lying entirely above the x-axis.

Try it yourself first. Then reveal the solution one step at a time.

2.3 Relationship between Zeroes and Coefficients of a Polynomial

Section 2.1 left us with a question: do the zeroes of a quadratic depend on its coefficients the way $- \frac{b}{a}$ does for a line ((2.1))? Let us experiment. We factorise two quadratics, list their zeroes and look for a pattern.

First, $p (x) = 2 x^{2} + 2 x - 12.$ Every coefficient is even, so take out 2: $p (x) = 2 (x^{2} + x - 6),$ which factorises as $2 (x + 3) (x - 2) .$

The product is zero only when $x + 3 = 0$ or $x - 2 = 0,$ so the zeroes of $p (x)$ are $- 3$ and $2.$

Next, $q (x) = 4 x^{2} - 4 x - 3.$ We split the middle term: find two numbers whose product is $a \times c = 4 \times (- 3) = - 12$ and whose sum is $b = - 4.$ They are $- 6$ and $2,$ so $- 4 x$ becomes $- 6 x + 2 x$ and $q (x) = 4 x^{2} - 6 x + 2 x - 3.$ Group the first two terms and the last two: $2 x (2 x - 3) + 1 (2 x - 3),$ which is $(2 x + 1) (2 x - 3) .$

So the zeroes of $q (x)$ are $- \frac{1}{2}$ and $\frac{3}{2} .$

Try this

Spot the pattern

Predict first

For

q (x) = 4 x^{2} - 4 x - 3

the sum of the zeroes is

- \frac{1}{2} + \frac{3}{2} = 1.

Which expression in

a = 4,

b = - 4,

c = - 3

also equals 1?

\frac{b}{a}

- \frac{b}{a}

\frac{c}{a}

- \frac{c}{b}

Fill in the last two columns for both polynomials. Then compare them with the sum and product columns.

Table 2.4Zeroes and coefficients

Polynomial	Zeroes	Sum	Product	$- \frac{b}{a}$	$\frac{c}{a}$
$2 x^{2} + 2 x - 12$	$- 3,$ $2$	$- 1$	$- 6$
$4 x^{2} - 4 x - 3$	$- \frac{1}{2},$ $\frac{3}{2}$	$1$	$- \frac{3}{4}$

Your entries are saved automatically.

Two examples could be a coincidence. Here is why the pattern holds for every quadratic that has zeroes.

DerivationSum and product of the zeroes of a quadratic

Suppose the quadratic $p (x) = a x^{2} + b x + c,$ with $a \neq 0,$ has zeroes $α$ and $β$ (Greek alpha and beta). Each zero gives a factor, by the Factor Theorem: $x - α$ and $x - β .$ The degree is 2, so nothing is left over except a constant multiplier $k :$ $p (x) = k (x - α) (x - β) .$

Multiplying out the brackets gives $p (x) = k x^{2} - k (α + β) x + k α β .$

Two polynomials are equal only when their matching coefficients are equal. Matching the $x^{2}$ terms, the $x$ terms and the constants:

\begin{aligned} a & = k \\ b & = - k (α + β) \\ c & = k α β \end{aligned}

Since $k = a,$ dividing by $a$ (which is not zero) gives the two relations.

Sum of the zeroes

α + β = - \frac{b}{a}

(2.2)

Product of the zeroes

α β = \frac{c}{a}

(2.3)

In words: the sum of the zeroes is minus the coefficient of $x$ divided by the coefficient of $x^{2} .$ The product of the zeroes is the constant term divided by the coefficient of $x^{2} .$

Tick Compare the zeroes with the coefficients in the Zero explorer (Fig. 2.7) to test the relations on any quadratic you like.

Example 2NCERT Example 2 · p. 20

Zeroes first, then the check

Find the zeroes of $5 t - 3 + 2 t^{2} .$ Then check that their sum and product match the coefficients.

Try it yourself first. Then reveal the solution one step at a time.

Example 3NCERT Example 3 · p. 20

Zeroes that are surds

Find the zeroes of $2 x^{2} - 14.$ Then check that their sum and product match the coefficients.

Try it yourself first. Then reveal the solution one step at a time.

The relations also work backwards. Suppose you know the sum $S$ and the product $P$ of the zeroes. Taking $a = 1,$ the relations give $b = - S$ and $c = P,$ so one quadratic with these zeroes is

Quadratic from the sum and product of its zeroes

x^{2} - S x + P

(2.4)

Multiplying by any non-zero constant $k$ does not change the zeroes. So every quadratic with these zeroes has the form $k (x^{2} - S x + P) .$ In Fig. 2.11, changing $k$ stretches or flips the parabola, but it still passes through the same two points of the $x$ -axis.

3/3 $y = - x^{2} + x + 2$ opens downward, with the same zeroes.

Fig. 2.11Same zeroes, many quadratics:

k (x^{2} - x - 2)

has zeroes

- 1

and

2

for every

k \neq 0.

A grid with x from minus 3 to 4 and y from minus 5 to 6, with the two points (minus 1, 0) and (2, 0) ringed throughout. In the first step the solid parabola y equals x squared minus x minus 2, for k equal to 1, passes through both rings. In the second step the dashed parabola for k equal to 2 appears; it is narrower and dips twice as deep, through the same rings. In the third step the dotted parabola for k equal to minus 1 appears; it opens downward, again through the same two rings.

Example 4NCERT Example 4 · p. 21

Build a quadratic from its zeroes' sum and product

The zeroes of a quadratic add up to $\frac{7}{2}$ and multiply to $\frac{3}{2} .$ Write down such a quadratic with whole-number coefficients.

Try it yourself first. Then reveal the solution one step at a time.

Example

Board-exam style: use the relations without solving

If $α$ and $β$ are the zeroes of $x^{2} - 6 x + 4,$ find (a) $α^{2} + β^{2}$ and (b) $\frac{1}{α} + \frac{1}{β},$ without finding $α$ and $β .$

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 4

What is the sum of the zeroes of

3 x^{2} - 6 x + 2

2

- 2

\frac{2}{3}

- 6

Show answer

Answer (A) $2$

- \frac{b}{a} = - \frac{- 6}{3} = 2.

Forgetting the minus sign gives

- 2,

and

\frac{2}{3}

\frac{c}{a},

the product.

Zeroes and coefficients of a cubic

Does a cubic follow a similar rule? Take $p (x) = 2 x^{3} + 3 x^{2} - 8 x + 3.$ Substituting shows that $1,$ $- 3$ and $\frac{1}{2}$ are zeroes:

$p (1) = 2 + 3 - 8 + 3 = 0$
$p (- 3) = - 54 + 27 + 24 + 3 = 0$
$p (\frac{1}{2}) = \frac{1}{4} + \frac{3}{4} - 4 + 3 = 0$

There is no room for a fourth zero, since a cubic has at most three. Table 2.5 compares their combinations with $a = 2,$ $b = 3,$ $c = - 8$ and $d = 3.$

Table 2.5The zeroes

1,

- 3,

\frac{1}{2}

2 x^{3} + 3 x^{2} - 8 x + 3

Combination	From the zeroes	From the coefficients
sum of the zeroes	$1 - 3 + \frac{1}{2} = - \frac{3}{2}$	$- \frac{b}{a} = - \frac{3}{2}$
sum of the products in pairs	$- 3 - \frac{3}{2} + \frac{1}{2} = - 4$	$\frac{c}{a} = \frac{- 8}{2} = - 4$
product of the zeroes	$1 \times (- 3) \times \frac{1}{2} = - \frac{3}{2}$	$- \frac{d}{a} = - \frac{3}{2}$

So a cubic has three relations, not two. The middle one is new: multiply the zeroes in pairs, $α β,$ $β γ$ and $γ α,$ and add the three results. Here is why the relations hold for every cubic.

DerivationRelations for a cubic

Let $α,$ $β,$ $γ$ (gamma) be the zeroes of $p (x) = a x^{3} + b x^{2} + c x + d .$ As before, the Factor Theorem gives $p (x) = a (x - α) (x - β) (x - γ) .$

Multiply out the brackets and collect the powers of $x .$ The product $(x - α) (x - β) (x - γ)$ equals $x^{3} - (α + β + γ) x^{2} + (α β + β γ + γ α) x - α β γ .$

Multiply by $a$ and compare with $a x^{3} + b x^{2} + c x + d .$ Then $b = - a (α + β + γ),$ $c = a (α β + β γ + γ α)$ and $d = - a α β γ .$ Dividing each by $a$ gives the three relations below.

Zeroes and coefficients of a cubic

\begin{aligned} α + β + γ & = - \frac{b}{a} \\ α β + β γ + γ α & = \frac{c}{a} \\ α β γ & = - \frac{d}{a} \end{aligned}

(2.5)

The signs alternate as you move from $b$ to $c$ to $d :$ minus, plus, minus.

Example 5NCERT Example 5 · p. 22

Check the zeroes of a cubic

Show by substitution that $2,$ $- 1$ and $- \frac{2}{3}$ are zeroes of $p (x) = 3 x^{3} - x^{2} - 8 x - 4.$ Then check all three zero–coefficient relations for it.

NCERT marks Example 5 as outside the exam. It is still the best practice for the cubic relations.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

What is the product of the zeroes of

x^{3} - 4 x^{2} + x + 6

Show answer

Answer -6

- \frac{d}{a} = - \frac{6}{1} = - 6.

The zeroes are

- 1,

2

and

3,

and

(- 1) (2) (3) = - 6.

Using

\frac{d}{a},

as for a quadratic's product, gives the wrong sign.

Practice · 10 questions

Practice questions with answers (NCERT “Practice 2.2”)

1
Short answerEasyNCERT Exercise 2.2, Q1
Factorise each quadratic polynomial to find its zeroes. Then check that their sum and product agree with the coefficients.
1. $x^{2} - 3 x - 28$
  Show solution
  Answer $7$ and $- 4$
  Two numbers with product $- 28$ and sum $- 3$ are $- 7$ and $4.$ So the polynomial is $(x - 7) (x + 4),$ with zeroes $7$ and $- 4.$
  Sum: $3 = - \frac{- 3}{1} .$ Product: $- 28 = \frac{- 28}{1} .$
2. $9 y^{2} - 12 y + 4$
  Show solution
  Answer $\frac{2}{3}$ and $\frac{2}{3}$ (equal zeroes)
  It is a perfect square, $(3 y - 2)^{2} .$ Both zeroes are $\frac{2}{3} .$
  Sum: $\frac{4}{3} = - \frac{- 12}{9} .$ Product: $\frac{4}{9} = \frac{4}{9} .$
3. $7 x - 12 + 10 x^{2}$
  Show solution
  Answer $\frac{4}{5}$ and $- \frac{3}{2}$
  Standard form: $10 x^{2} + 7 x - 12.$ Here $a c = - 120;$ $15$ and $- 8$ multiply to $- 120$ and add to $7.$
  Grouping gives $(5 x - 4) (2 x + 3),$ so the zeroes are $\frac{4}{5}$ and $- \frac{3}{2} .$
  Sum: $- \frac{7}{10} = - \frac{b}{a} .$ Product: $- \frac{6}{5} = \frac{- 12}{10} .$
4. $3 v^{2} - 12 v$
  Show solution
  Answer $0$ and $4$
  Take out the common factor $3 v$ to get $3 v (v - 4) .$ The zeroes are $0$ and $4.$
  Here $c = 0.$ Sum: $4 = - \frac{- 12}{3} .$ Product: $0 = \frac{0}{3} .$
2
Short answerMediumNCERT Exercise 2.2, Q1
Find the zeroes, then check their sum and product against the coefficients.
1. $p^{2} - 11$
  Show solution
  Answer $\sqrt{11}$ and $- \sqrt{11}$
  Write $11$ as $(\sqrt{11})^{2}$ and use the difference of squares. The zeroes are $\sqrt{11}$ and $- \sqrt{11} .$
  There is no $p$ term, so $b = 0.$ Sum: $0 = - \frac{0}{1} .$ Product: $- 11 = \frac{- 11}{1} .$
2. $5 x^{2} + 4 x - 1$
  Show solution
  Answer $\frac{1}{5}$ and $- 1$
  Here $a c = - 5;$ $5$ and $- 1$ multiply to $- 5$ and add to $4.$ So $5 x^{2} + 5 x - x - 1$ groups as $(5 x - 1) (x + 1) .$
  Zeroes: $\frac{1}{5}$ and $- 1.$ Sum: $- \frac{4}{5} = - \frac{b}{a} .$ Product: $- \frac{1}{5} = \frac{c}{a} .$
3
Short answerEasyNCERT Exercise 2.2, Q2
Find a quadratic polynomial whose zeroes have the given sum and product.
1. Sum $- 2,$ product $- 15$
  Show solution
  Answer $x^{2} + 2 x - 15$
  $S = - 2$ and $P = - 15,$ so $x^{2} - S x + P$ becomes $x^{2} + 2 x - 15.$
2. Sum $\frac{1}{3},$ product $- 2$
  Show solution
  Answer $3 x^{2} - x - 6$
  $x^{2} - \frac{1}{3} x - 2.$ Multiply by $3$ to get $3 x^{2} - x - 6.$
3. Sum $0,$ product $- \frac{3}{4}$
  Show solution
  Answer $4 x^{2} - 3$
  $x^{2} - 0 x - \frac{3}{4} .$ Multiply by $4$ to get $4 x^{2} - 3,$ whose zeroes are $\pm \frac{\sqrt{3}}{2} .$
4. Sum $2 \sqrt{3},$ product $3$
  Show solution
  Answer $x^{2} - 2 \sqrt{3} x + 3$
  $x^{2} - 2 \sqrt{3} x + 3.$ It equals $(x - \sqrt{3})^{2},$ so both zeroes are $\sqrt{3} .$
4
Short answerMediumNCERT Exercise 2.2, Q2
Find a quadratic polynomial whose zeroes have the given sum and product.
1. Sum $- \frac{3}{5},$ product $0$
  Show solution
  Answer $5 x^{2} + 3 x$
  $x^{2} + \frac{3}{5} x + 0.$ Multiply by $5$ to get $5 x^{2} + 3 x,$ with zeroes $0$ and $- \frac{3}{5} .$
2. Sum $- \frac{1}{2},$ product $- \frac{1}{2}$
  Show solution
  Answer $2 x^{2} + x - 1$
  $x^{2} + \frac{1}{2} x - \frac{1}{2} .$ Multiply by $2$ to get $2 x^{2} + x - 1,$ which is $(2 x - 1) (x + 1) .$
5
Short answerMedium
If $α$ and $β$ are the zeroes of $3 x^{2} - 6 x + 1,$ find each value without finding $α$ and $β .$
1. $\frac{1}{α} + \frac{1}{β}$
  Show answer
  Answer 6
  $α + β = 2$ and $α β = \frac{1}{3} .$ So $\frac{α + β}{α β} = 2 \div \frac{1}{3} = 6.$
2. $α^{2} β + α β^{2}$
  Show answer
  Answer 0.6667
  Factorise: $α β (α + β) = \frac{1}{3} \times 2 = \frac{2}{3} .$
3. $(α - β)^{2}$
  Show answer
  Answer 2.6667
  $(α - β)^{2}$ equals $(α + β)^{2} - 4 α β,$ which is $4 - \frac{4}{3} = \frac{8}{3} .$
Quick answerHard
One zero of $x^{2} - 9 x + k$ is twice the other. Find $k .$
Show answer
Answer 18
The sum is $m + 2 m = 3 m = 9,$ so $m = 3$ and the zeroes are $3$ and $6.$ Then $k$ is the product, $18.$
Quick answerMedium
Without solving, what can you say about the zeroes of $x^{2} + 17 x + 41$ ? (It does have two zeroes.)
both positiveboth negativeone positive, one negativeboth equal
Show answer
Answer (B) both negative
The product $41$ is positive, so the zeroes have the same sign. Their sum $- 17$ is negative, so both are negative. (They are about $- 2.9$ and $- 14.1$ .)
8
Short answerMedium
For the cubic polynomial $2 x^{3} - 7 x^{2} + 2 x + 3 :$
1. Find the sum of its zeroes.
  Show answer
  Answer 3.5
  $- \frac{b}{a} = - \frac{- 7}{2} = \frac{7}{2} .$
2. Find the sum of the products of its zeroes taken two at a time.
  Show answer
  Answer 1
  $\frac{c}{a} = \frac{2}{2} = 1.$
3. Find the product of its zeroes.
  Show answer
  Answer -1.5
  $- \frac{d}{a} = - \frac{3}{2} .$
4. Two of its zeroes are $3$ and $1.$ Use (a) to find the third.
  Show answer
  Answer -0.5
  $3 + 1 + γ = \frac{7}{2},$ so $γ = - \frac{1}{2} .$ Check with (c): $3 \times 1 \times (- \frac{1}{2}) = - \frac{3}{2} .$
Quick answerMedium
Assertion (A): $x^{2} + 6$ has no zero. Reason (R): the graph of $y = x^{2} + 6$ is a parabola that opens upward, with its lowest point at $(0, 6) .$
Both A and R are true, and R explains A.Both A and R are true, but R does not explain A.A is true but R is false.A is false but R is true.
Show answer
Answer (A) Both A and R are true, and R explains A.
Both are true, since $x^{2} + 6 \geq 6$ for every $x .$ The lowest point of the parabola is above the $x$ -axis, so the graph never meets the axis. That is exactly why there is no zero.
Long answerHard
Find a quadratic polynomial whose zeroes are $2 + \sqrt{3}$ and $2 - \sqrt{3} .$ Then check your answer by finding its value at $2 + \sqrt{3} .$
Show solution
Answer $x^{2} - 4 x + 1$
Sum: the $\sqrt{3}$ terms cancel, so $S = 4.$
Product: by $m^{2} - n^{2},$ $P = 4 - 3 = 1.$
So the polynomial is $x^{2} - 4 x + 1$ (or any non-zero multiple of it).
Check: at $x = 2 + \sqrt{3},$ $x^{2} = 7 + 4 \sqrt{3}$ and $4 x = 8 + 4 \sqrt{3} .$ So the value is $7 - 8 + 1 = 0.$

2.4 Summary

Fig. 2.12 puts the whole chapter on one polynomial: its formula, its graph, and the link between its zeroes and its coefficients.

3/3Sum $- 2 + 3 = 1 = - \frac{b}{a};$ product $(- 2) (3) = - 6 = \frac{c}{a} .$

Fig. 2.12One polynomial, three views: formula, graph and the zero–coefficient relations.

On the left, a grid with the upward parabola y equals x squared minus x minus 6. In the first step the right side shows the formula and its coefficients a equals 1, b equals minus 1, c equals minus 6. In the second step the points where the parabola meets the x-axis, minus 2 and 3, are ringed and named. In the third step a shaded box shows that minus 2 plus 3 equals 1, which is minus b over a, and that minus 2 times 3 equals minus 6, which is c over a.

Recap

Polynomials: key points and summary

Linear, quadratic and cubic are the names for degrees 1, 2 and 3. A quadratic is $a x^{2} + b x + c$ and a cubic is $a x^{3} + b x^{2} + c x + d,$ with $a \neq 0$ (Section 2.1).
$k$ is a zero of $p (x)$ when $p (k) = 0.$ The zero of $a x + b$ is $- \frac{b}{a}$ (Value and zero of a polynomial).
Each zero of $p (x)$ is the $x$ -value of a point where the graph of $y = p (x)$ crosses or touches the $x$ -axis (Section 2.2).
The graph of a quadratic is a parabola. It opens up when $a > 0$ and down when $a < 0$ (Quadratic polynomials).
A quadratic has two zeroes, two equal zeroes or no zero (Two zeroes, one zero or none). A cubic has one, two or three (Cubic polynomials).
Degree $n$ allows no more than $n$ zeroes (Property).
For a quadratic with zeroes $α$ and $β :$ $α + β = - \frac{b}{a}$ and $α β = \frac{c}{a}$ (Derivation).
A quadratic whose zeroes have sum $S$ and product $P$ is $k (x^{2} - S x + P),$ with $k \neq 0$ ((2.4)).
For a cubic with zeroes $α,$ $β,$ $γ :$ the sum is $- \frac{b}{a},$ the sum of the products in pairs is $\frac{c}{a}$ and the product is $- \frac{d}{a}$ (Zeroes and coefficients of a cubic).

Flash cards · 8

Formula recall

1 / 8

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Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercise 2.1, Q1	Practice 2.1 · Q1
NCERT Exercise 2.2, Q1	Practice 2.2 · Q1
NCERT Exercise 2.2, Q1	Practice 2.2 · Q2
NCERT Exercise 2.2, Q2	Practice 2.2 · Q3
NCERT Exercise 2.2, Q2	Practice 2.2 · Q4

Quick check: MCQs on Polynomials

Value and zero of a polynomial4 questions
Cubic polynomials4 questions
2.3 Relationship between Zeroes and Coefficients of a Polynomial4 questions
Zeroes and coefficients of a cubic3 questions

Glossary · 8

Key terms (English–Hindi)

coefficientगुणांकgunaank: The number that multiplies a power of the variable in a term, such as $- 4$ in $- 4 x^{2}$ .First seen in the chapter →
cubic polynomialत्रिघात बहुपदtrighaat bahupad: A polynomial of degree 3, of the form $a x^{3} + b x^{2} + c x + d$ with $a \neq 0$ .First seen in the chapter →

degreeघातghaat: The highest power of the variable in a polynomial, such as 3 for $2 x^{3} - x + 5$ .First seen in the chapter →

Factor Theoremगुणनखंड प्रमेयgunankhand prameya: If $p (α) = 0$ , then $x - α$ is a factor of $p (x)$ , and the other way round.First seen in the chapter →

linear polynomialरैखिक बहुपदraikhik bahupad: A polynomial of degree 1, of the form $a x + b$ with $a \neq 0$ .First seen in the chapter →

parabolaपरवलयparvalay: The U-shaped graph of a quadratic polynomial. It opens upward when $a > 0$ and downward when $a < 0$ .First seen in the chapter →

quadratic polynomialद्विघात बहुपदdvighaat bahupad: A polynomial of degree 2, of the form $a x^{2} + b x + c$ with $a \neq 0$ .First seen in the chapter →

zero of a polynomialबहुपद का शून्यकbahupad ka shoonyak: A real number $k$ with $p (k) = 0$ . On the graph of $y = p (x)$ it is where the curve meets the $x$ -axis.

Polynomials: Class 10 Maths · Chapter 2

What you will learn in Polynomials

2.1 Introduction

Value and zero of a polynomial

2.2 Geometrical Meaning of the Zeroes of a Polynomial

Linear polynomials

Quadratic polynomials

Two zeroes, one zero or none

Cubic polynomials

Practice 2.1: questions with answers

2.3 Relationship between Zeroes and Coefficients of a Polynomial

Zeroes and coefficients of a cubic

Practice questions with answers (NCERT “Practice 2.2”)

2.4 Summary

Polynomials: key points and summary

Find your NCERT exercise question

Quick check: MCQs on Polynomials

Key terms (English–Hindi)

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