NCERT Solutions for Class 10 Maths Chapter 2: Polynomials — Complete Guide with Step-by-Step Solutions

A polynomial in variable $x$ is an expression of the form:

where $a_0, a_1, \ldots, a_n$ are real numbers and $n$ is a non-negative integer.

• The degree of a polynomial is the highest power of $x$ with a non-zero coefficient.
  - A polynomial of degree 1 is linear: $ax + b$ (one zero).
  - A polynomial of degree 2 is quadratic: $ax^2 + bx + c$ (at most two zeroes).
  - A polynomial of degree 3 is cubic: $ax^3 + bx^2 + cx + d$ (at most three zeroes).

The zeroes (or roots) of $p(x)$ are the values of $x$ for which $p(x) = 0$.

The zeroes of a polynomial $p(x)$ are the **x-coordinates of the points where the graph of $y = p(x)$ intersects the x-axis**.

• A linear polynomial $ax + b$ has exactly 1 zero, and its graph is a straight line crossing the x-axis at $x = -\frac{b}{a}$.
• A quadratic polynomial $ax^2 + bx + c$ has at most 2 zeroes, and its graph is a parabola. The parabola opens upward if $a > 0$ and downward if $a < 0$. It may cross the x-axis at 0, 1, or 2 points.
• A cubic polynomial has at most 3 zeroes, and its graph is an S-shaped curve that may cross the x-axis 1, 2, or 3 times.

The number of zeroes equals the number of x-axis intersections.

For a quadratic polynomial $p(x) = ax^2 + bx + c$ with zeroes $\alpha$ and $\beta$:

Conversely, if the zeroes are $\alpha$ and $\beta$, the quadratic polynomial is:

Memory aid: For the sum, think "negative $b$ over $a$" — the sign flips. For the product, think "$c$ over $a$" — no sign change.

For a cubic polynomial $p(x) = ax^3 + bx^2 + cx + d$ with zeroes $\alpha$, $\beta$, $\gamma$:

Notice the pattern of alternating signs: $-b/a$, $+c/a$, $-d/a$. This pattern (called Vieta's formulas) extends to higher degree polynomials.

Statement: Given any polynomial $p(x)$ and any non-zero polynomial $g(x)$, there exist polynomials $q(x)$ (quotient) and $r(x)$ (remainder) such that:

This is the polynomial version of Euclid's Division Lemma from Chapter 1. The process of finding $q(x)$ and $r(x)$ is called polynomial long division.

Key application: If $\alpha$ is a zero of $p(x)$, then $(x - \alpha)$ is a factor of $p(x)$, meaning the remainder when dividing by $(x - \alpha)$ is zero. This is the Factor Theorem.

Solution:
The number of zeroes = number of times the graph intersects (or touches) the x-axis.

(i) The graph is a straight line parallel to the x-axis (does not intersect x-axis).
$\Rightarrow$ Number of zeroes $= 0$.

(ii) The graph intersects the x-axis at exactly one point.
$\Rightarrow$ Number of zeroes $= 1$.

(iii) The graph intersects the x-axis at exactly two points.
$\Rightarrow$ Number of zeroes $= 2$.

(iv) The graph intersects the x-axis at exactly two points.
$\Rightarrow$ Number of zeroes $= 2$.

(v) The graph intersects the x-axis at exactly four points.
$\Rightarrow$ Number of zeroes $= 4$.

(vi) The graph intersects the x-axis at exactly three points.
$\Rightarrow$ Number of zeroes $= 3$.

Key insight: A polynomial of degree $n$ can have at most $n$ zeroes, but it can have fewer. For example, $x^2 + 1$ is a quadratic with zero real zeroes (its graph never crosses the x-axis).

Solution:
Factorise by splitting the middle term:

Zeroes: $x = 4$ and $x = -2$.

Verification:
Here $a = 1$, $b = -2$, $c = -8$.

Sum of zeroes $= 4 + (-2) = 2$
$-\frac{b}{a} = -\frac{-2}{1} = 2$ $\checkmark$

Product of zeroes $= 4 \times (-2) = -8$
$\frac{c}{a} = \frac{-8}{1} = -8$ $\checkmark$

Solution:

This is a perfect square! Zeroes: $s = \frac{1}{2}$ (repeated).

Verification:
Here $a = 4$, $b = -4$, $c = 1$.

Sum of zeroes $= \frac{1}{2} + \frac{1}{2} = 1$
$-\frac{b}{a} = -\frac{-4}{4} = 1$ $\checkmark$

Product of zeroes $= \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
$\frac{c}{a} = \frac{1}{4}$ $\checkmark$

Note: When a quadratic has a repeated zero, its graph touches the x-axis at exactly one point (the vertex lies on the x-axis).

Solution:
First, rewrite in standard form: $6x^2 - 7x - 3$.

Split the middle term. We need two numbers that multiply to $6 \times (-3) = -18$ and add to $-7$. Those are $-9$ and $+2$.

Zeroes: $x = -\frac{1}{3}$ and $x = \frac{3}{2}$.

Verification:
Here $a = 6$, $b = -7$, $c = -3$.

Sum $= -\frac{1}{3} + \frac{3}{2} = \frac{-2 + 9}{6} = \frac{7}{6}$
$-\frac{b}{a} = -\frac{-7}{6} = \frac{7}{6}$ $\checkmark$

Product $= \left(-\frac{1}{3}\right) \times \frac{3}{2} = -\frac{1}{2} = -\frac{3}{6}$
$\frac{c}{a} = \frac{-3}{6} = -\frac{1}{2}$ $\checkmark$

Solution:

Zeroes: $u = 0$ and $u = -2$.

Verification:
Here $a = 4$, $b = 8$, $c = 0$.

Sum $= 0 + (-2) = -2$
$-\frac{b}{a} = -\frac{8}{4} = -2$ $\checkmark$

Product $= 0 \times (-2) = 0$
$\frac{c}{a} = \frac{0}{4} = 0$ $\checkmark$

Note: Zero itself can be a zero of a polynomial! When $c = 0$, the polynomial has $x = 0$ as one of its zeroes.

Solution:

Zeroes: $t = \sqrt{15}$ and $t = -\sqrt{15}$.

Verification:
Here $a = 1$, $b = 0$, $c = -15$.

Sum $= \sqrt{15} + (-\sqrt{15}) = 0$
$-\frac{b}{a} = -\frac{0}{1} = 0$ $\checkmark$

Product $= \sqrt{15} \times (-\sqrt{15}) = -15$
$\frac{c}{a} = \frac{-15}{1} = -15$ $\checkmark$

Solution:
Split: we need two numbers multiplying to $3 \times (-4) = -12$ and adding to $-1$. Those are $-4$ and $+3$.

Zeroes: $x = -1$ and $x = \frac{4}{3}$.

Verification:
$a = 3$, $b = -1$, $c = -4$.

Sum $= -1 + \frac{4}{3} = \frac{-3 + 4}{3} = \frac{1}{3}$
$-\frac{b}{a} = -\frac{-1}{3} = \frac{1}{3}$ $\checkmark$

Product $= (-1) \times \frac{4}{3} = -\frac{4}{3}$
$\frac{c}{a} = \frac{-4}{3}$ $\checkmark$

Formula: If sum of zeroes $= S$ and product of zeroes $= P$, the quadratic polynomial is:

(i) $S = \frac{1}{4}$, $P = -1$:

(ii) $S = \sqrt{2}$, $P = \frac{1}{3}$:

(iii) $S = 0$, $P = \sqrt{5}$:

(iv) $S = 1$, $P = 1$:

(v) $S = -\frac{1}{4}$, $P = \frac{1}{4}$:

(vi) $S = 4$, $P = 1$:

Solution:
We perform polynomial long division.

Step 1: Divide the leading term: $\frac{x^3}{x^2} = x$. Write $x$ as the first term of the quotient.

$x \times (x^2 - 2) = x^3 - 2x$

Subtract from $p(x)$:
$(x^3 - 3x^2 + 5x - 3) - (x^3 - 2x) = -3x^2 + 7x - 3$

Step 2: Divide: $\frac{-3x^2}{x^2} = -3$.

$-3 \times (x^2 - 2) = -3x^2 + 6$

Subtract:
$(-3x^2 + 7x - 3) - (-3x^2 + 6) = 7x - 9$

Since $\deg(7x - 9) = 1 < 2 = \deg(g)$, we stop.

Quotient: $q(x) = x - 3$
Remainder: $r(x) = 7x - 9$

Verification: $(x^2 - 2)(x - 3) + (7x - 9) = x^3 - 3x^2 - 2x + 6 + 7x - 9 = x^3 - 3x^2 + 5x - 3 = p(x)$ $\checkmark$

Solution:
Rewrite $g(x) = x^2 - x + 1$.

Step 1: $\frac{x^4}{x^2} = x^2$.
$x^2(x^2 - x + 1) = x^4 - x^3 + x^2$
Subtract: $(x^4 - 3x^2 + 4x + 5) - (x^4 - x^3 + x^2) = x^3 - 4x^2 + 4x + 5$

Step 2: $\frac{x^3}{x^2} = x$.
$x(x^2 - x + 1) = x^3 - x^2 + x$
Subtract: $(x^3 - 4x^2 + 4x + 5) - (x^3 - x^2 + x) = -3x^2 + 3x + 5$

Step 3: $\frac{-3x^2}{x^2} = -3$.
$-3(x^2 - x + 1) = -3x^2 + 3x - 3$
Subtract: $(-3x^2 + 3x + 5) - (-3x^2 + 3x - 3) = 8$

Quotient: $q(x) = x^2 + x - 3$
Remainder: $r(x) = 8$

Verification: $(x^2 - x + 1)(x^2 + x - 3) + 8 = x^4 - 3x^2 + 4x + 5$ $\checkmark$ (expand to verify)

Solution:
Rewrite: $g(x) = -x^2 + 2$. For convenience, we can also write $g(x) = -(x^2 - 2)$.

Note: $p(x) = x^4 + 0x^3 + 0x^2 - 5x + 6$.

Divide by $-x^2 + 2$:

Step 1: $\frac{x^4}{-x^2} = -x^2$.
$-x^2(-x^2 + 2) = x^4 - 2x^2$
Subtract: $(x^4 - 5x + 6) - (x^4 - 2x^2) = 2x^2 - 5x + 6$

Step 2: $\frac{2x^2}{-x^2} = -2$.
$-2(-x^2 + 2) = 2x^2 - 4$
Subtract: $(2x^2 - 5x + 6) - (2x^2 - 4) = -5x + 10$

Quotient: $q(x) = -x^2 - 2$
Remainder: $r(x) = -5x + 10$

Verification: $(-x^2 + 2)(-x^2 - 2) + (-5x + 10) = (x^4 - 4) + (-5x + 10) = x^4 - 5x + 6$ $\checkmark$

**$g(x)$ is a factor of $p(x)$ if and only if the remainder is zero when dividing $p(x)$ by $g(x)$.**

(i) $p(x) = 2x^3 + x^2 - 2x - 1$, $g(x) = x + 1$.

By the Factor Theorem, $(x + 1)$ is a factor iff $p(-1) = 0$.
$p(-1) = 2(-1)^3 + (-1)^2 - 2(-1) - 1 = -2 + 1 + 2 - 1 = 0$

Yes, $g(x)$ is a factor. $\checkmark$

Actually, $2x^3 + x^2 - 2x - 1 = x^2(2x + 1) - (2x + 1) = (x^2 - 1)(2x + 1) = (x-1)(x+1)(2x+1)$.

(ii) $p(x) = x^3 + 3x + 1$, $g(x) = x^2 + 1$.

Divide: $\frac{x^3}{x^2} = x$. $x(x^2 + 1) = x^3 + x$.
Subtract: $(x^3 + 3x + 1) - (x^3 + x) = 2x + 1$.

Remainder $= 2x + 1 \neq 0$.

No, $g(x)$ is not a factor.

(iii) $p(x) = x^3 - 4x^2 + x + 6$, $g(x) = x - 3$.

Check $p(3) = 27 - 36 + 3 + 6 = 0$.

Yes, $g(x)$ is a factor. $\checkmark$

Full factorisation: $x^3 - 4x^2 + x + 6 = (x - 3)(x^2 - x - 2) = (x-3)(x-2)(x+1)$.

Solution:
Since $\sqrt{2}$ and $-\sqrt{2}$ are zeroes, $(x - \sqrt{2})(x + \sqrt{2}) = x^2 - 2$ is a factor.

Divide $2x^4 - 3x^3 - 3x^2 + 6x - 2$ by $x^2 - 2$:

Step 1: $\frac{2x^4}{x^2} = 2x^2$.
$2x^2(x^2 - 2) = 2x^4 - 4x^2$
Subtract: $(2x^4 - 3x^3 - 3x^2 + 6x - 2) - (2x^4 - 4x^2) = -3x^3 + x^2 + 6x - 2$

Step 2: $\frac{-3x^3}{x^2} = -3x$.
$-3x(x^2 - 2) = -3x^3 + 6x$
Subtract: $(-3x^3 + x^2 + 6x - 2) - (-3x^3 + 6x) = x^2 - 2$

Step 3: $\frac{x^2}{x^2} = 1$.
$1(x^2 - 2) = x^2 - 2$
Subtract: $(x^2 - 2) - (x^2 - 2) = 0$

Quotient $= 2x^2 - 3x + 1$. Factorise:

Remaining zeroes: $x = \frac{1}{2}$ and $x = 1$.

All four zeroes: $\sqrt{2}, -\sqrt{2}, \frac{1}{2}, 1$

Complete factorisation: $2x^4 - 3x^3 - 3x^2 + 6x - 2 = (x^2 - 2)(2x - 1)(x - 1)$

Solution:
The factor from the two given zeroes:

Divide $x^4 - 6x^3 - 26x^2 + 138x - 35$ by $x^2 - 4x + 1$:

Step 1: $\frac{x^4}{x^2} = x^2$.
$x^2(x^2 - 4x + 1) = x^4 - 4x^3 + x^2$
Subtract: remainder $= -2x^3 - 27x^2 + 138x - 35$

Step 2: $\frac{-2x^3}{x^2} = -2x$.
$-2x(x^2 - 4x + 1) = -2x^3 + 8x^2 - 2x$
Subtract: remainder $= -35x^2 + 140x - 35$

Step 3: $\frac{-35x^2}{x^2} = -35$.
$-35(x^2 - 4x + 1) = -35x^2 + 140x - 35$
Subtract: remainder $= 0$

Quotient $= x^2 - 2x - 35$. Factorise:

All four zeroes: $2 + \sqrt{3}, 2 - \sqrt{3}, 7, -5$

Solution:
Rearranging in standard form:
$p(x) = -x^3 + 3x^2 - 3x + 5$
$g(x) = -x^2 + x - 1$

Step 1: $\frac{-x^3}{-x^2} = x$.
$x(-x^2 + x - 1) = -x^3 + x^2 - x$
Subtract: $(-x^3 + 3x^2 - 3x + 5) - (-x^3 + x^2 - x) = 2x^2 - 2x + 5$

Step 2: $\frac{2x^2}{-x^2} = -2$.
$-2(-x^2 + x - 1) = 2x^2 - 2x + 2$
Subtract: $(2x^2 - 2x + 5) - (2x^2 - 2x + 2) = 3$

Quotient: $q(x) = x - 2$
Remainder: $r(x) = 3$

Verification: $g(x) \times q(x) + r(x) = (-x^2 + x - 1)(x - 2) + 3$
$= -x^3 + 2x^2 + x^2 - 2x - x + 2 + 3 = -x^3 + 3x^2 - 3x + 5 = p(x)$ $\checkmark$

Problem: Verify the relationship between zeroes and coefficients for $2x^3 + x^2 - 5x + 2$, given that its zeroes are $\frac{1}{2}$, $1$, and $-2$.

Solution:
Here $a = 2$, $b = 1$, $c = -5$, $d = 2$. Let $\alpha = \frac{1}{2}$, $\beta = 1$, $\gamma = -2$.

Sum of zeroes:
$\alpha + \beta + \gamma = \frac{1}{2} + 1 + (-2) = -\frac{1}{2}$
$-\frac{b}{a} = -\frac{1}{2}$ $\checkmark$

Sum of products taken two at a time:
$\alpha\beta + \beta\gamma + \gamma\alpha = \frac{1}{2}(1) + 1(-2) + (-2)\frac{1}{2} = \frac{1}{2} - 2 - 1 = -\frac{5}{2}$
$\frac{c}{a} = \frac{-5}{2}$ $\checkmark$

Product of zeroes:
$\alpha\beta\gamma = \frac{1}{2} \times 1 \times (-2) = -1$
$-\frac{d}{a} = -\frac{2}{2} = -1$ $\checkmark$

Solution:
Sum $= 3 + (-3) + 1 = 1$
Sum of products in pairs $= 3(-3) + (-3)(1) + (1)(3) = -9 - 3 + 3 = -9$
Product $= 3 \times (-3) \times 1 = -9$

The cubic polynomial is:

Verification: $p(3) = 27 - 9 - 27 + 9 = 0$ $\checkmark$, $p(-3) = -27 - 9 + 27 + 9 = 0$ $\checkmark$, $p(1) = 1 - 1 - 9 + 9 = 0$ $\checkmark$

Solution:
Let the zeroes be $\alpha$ and $7\alpha$.

Sum: $\alpha + 7\alpha = 8\alpha = \frac{8}{3}$, so $\alpha = \frac{1}{3}$.

Product: $\alpha \times 7\alpha = 7\alpha^2 = \frac{2k+1}{3}$.

$7 \times \frac{1}{9} = \frac{2k+1}{3}$

$\frac{7}{9} = \frac{2k+1}{3}$

$2k + 1 = \frac{7}{3}$

$2k = \frac{7}{3} - 1 = \frac{4}{3}$

$k = \frac{2}{3}$

Answer: $k = \frac{2}{3}$

Solution:
From the polynomial ($p = 1$, $q = -6$, $r = a$):
$\alpha + \beta = 6 \quad \cdots (1)$
$\alpha\beta = a \quad \cdots (2)$

Given: $3\alpha + 2\beta = 20 \quad \cdots (3)$

From (1): $\beta = 6 - \alpha$. Substitute in (3):
$3\alpha + 2(6 - \alpha) = 20$
$3\alpha + 12 - 2\alpha = 20$
$\alpha = 8$

So $\beta = 6 - 8 = -2$.

From (2): $a = \alpha\beta = 8 \times (-2) = -16$.

Answer: $a = -16$

Solution:
$x^2 + 7x + 10 = (x + 5)(x + 2)$

Zeroes: $x = -5$ and $x = -2$.

Verification: Sum $= -5 + (-2) = -7 = -\frac{7}{1} = -\frac{b}{a}$ $\checkmark$
Product $= (-5)(-2) = 10 = \frac{10}{1} = \frac{c}{a}$ $\checkmark$

Solution:
$x^2 - 3 = (x - \sqrt{3})(x + \sqrt{3})$

Zeroes: $x = \sqrt{3}$ and $x = -\sqrt{3}$.

Verification: Sum $= \sqrt{3} + (-\sqrt{3}) = 0 = -\frac{0}{1}$ $\checkmark$
Product $= \sqrt{3} \times (-\sqrt{3}) = -3 = \frac{-3}{1}$ $\checkmark$

Solution:
Sum $= (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4$
Product $= (2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1$

Observation: When zeroes are conjugate surds ($a \pm \sqrt{b}$), the sum and product are always rational, so the polynomial has rational coefficients.

Solution:
$\alpha + \beta = -\frac{5}{2}$ and $\alpha\beta = \frac{k}{2}$.

We know that $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$.

$\frac{21}{4} = \left(-\frac{5}{2}\right)^2 - 2 \cdot \frac{k}{2}$

$\frac{21}{4} = \frac{25}{4} - k$

$k = \frac{25}{4} - \frac{21}{4} = 1$

Answer: $k = 1$

Solution:
Let the zeroes be $\alpha$ and $\frac{1}{\alpha}$.

Product of zeroes $= \alpha \times \frac{1}{\alpha} = 1$.

But product $= \frac{c}{a} = \frac{k}{5}$.

So $\frac{k}{5} = 1 \implies k = 5$.

Answer: $k = 5$