NCERT Solutions for Class 10 Maths Chapter 2 – Polynomials – Exercise 2.3

Q1

Divide the polynomial $p(x)$ by the polynomial $g(x)$ and find the quotient and remainder in each of the following:

(i) $p(x) = x^3 – 3x^2 + 5x – 3$, $g(x) = x^2 – 2$

(ii) $p(x) = x^4 – 3x^2 + 4x + 5$, $g(x) = x^2 + 1 – x$

(iii) $p(x) = x^4 – 5x + 6$, $g(x) = 2 – x^2$

(i) $p(x) = x^3 – 3x^2 + 5x – 3$ divided by $g(x) = x^2 – 2$

x – 3 ________________ x² – 2 | x³ – 3x² + 5x – 3 x³ – 2x ________________ – 3x² + 7x – 3 – 3x² + 6 ________________ 7x – 9

✔️ Quotient = $x – 3$ | Remainder = $7x – 9$

(ii) $p(x) = x^4 – 3x^2 + 4x + 5$ divided by $g(x) = x^2 – x + 1$

Rearranging $g(x) = x^2 – x + 1$

x² + x – 3 ____________________ x²-x+1 | x⁴ + 0x³ – 3x² + 4x + 5 x⁴ – x³ + x² ____________________ x³ – 4x² + 4x x³ – x² + x ____________________ – 3x² + 3x + 5 – 3x² + 3x – 3 ____________________ 8

✔️ Quotient = $x^2 + x – 3$ | Remainder = $8$

(iii) $p(x) = x^4 – 5x + 6$ divided by $g(x) = 2 – x^2$ or $-x^2 + 2$

-x² – 2 ____________________ -x²+2 | x⁴ + 0x³ + 0x² – 5x + 6 x⁴ – 2x² ____________________ 2x² – 5x + 6 2x² – 4 ____________________ – 5x + 10

✔️ Quotient = $-x^2 – 2$ | Remainder = $-5x + 10$

Q2

Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

(i) $t^2 – 3$, $2t^4 + 3t^3 – 2t^2 – 9t – 12$

(ii) $x^2 + 3x + 1$, $3x^4 + 5x^3 – 7x^2 + 2x + 2$

(iii) $x^3 – 3x + 1$, $x^5 – 4x^3 + x^2 + 3x + 1$

(i) Divide $2t^4 + 3t^3 – 2t^2 – 9t – 12$ by $t^2 – 3$

2t² + 3t + 4 ____________________ t² – 3 | 2t⁴ + 3t³ – 2t² – 9t – 12 2t⁴ – 6t² ____________________ 3t³ + 4t² – 9t 3t³ – 9t ____________________ 4t² – 12 4t² – 12 ____________________ 0

Since the remainder is 0:

✓ Yes, $t^2 – 3$ is a factor of $2t^4 + 3t^3 – 2t^2 – 9t – 12$

(ii) Divide $3x^4 + 5x^3 – 7x^2 + 2x + 2$ by $x^2 + 3x + 1$

3x² – 4x + 2 ____________________ x²+3x+1 | 3x⁴ + 5x³ – 7x² + 2x + 2 3x⁴ + 9x³ + 3x² ____________________ – 4x³ – 10x² + 2x – 4x³ – 12x² – 4x ____________________ 2x² + 6x + 2 2x² + 6x + 2 ____________________ 0

Since the remainder is 0:

✓ Yes, $x^2 + 3x + 1$ is a factor of $3x^4 + 5x^3 – 7x^2 + 2x + 2$

(iii) Divide $x^5 – 4x^3 + x^2 + 3x + 1$ by $x^3 – 3x + 1$

x² – 1 ____________________ x³-3x+1 | x⁵ + 0x⁴ – 4x³ + x² + 3x + 1 x⁵ – 3x³ + x² ____________________ – x³ + 3x + 1 – x³ + 3x – 1 ____________________ 2

Since the remainder is 2 $\neq$ 0:

✖️ No, $x^3 – 3x + 1$ is NOT a factor of $x^5 – 4x^3 + x^2 + 3x + 1$

Q3

Obtain all other zeroes of $3x^4 + 6x^3 – 2x^2 – 10x – 5$, if two of its zeroes are $\sqrt{\frac{5}{3}}$ and $-\sqrt{\frac{5}{3}}$.

Solution

Since two zeroes are $\sqrt{\frac{5}{3}}$ and $-\sqrt{\frac{5}{3}}$, then:

$$\left(x – \sqrt{\frac{5}{3}}\right)\left(x + \sqrt{\frac{5}{3}}\right) = x^2 – \frac{5}{3}$$

So, $3(x^2 – \frac{5}{3}) = 3x^2 – 5$ is a factor of the given polynomial.

Dividing $3x^4 + 6x^3 – 2x^2 – 10x – 5$ by $3x^2 – 5$:

x² + 2x + 1 ____________________ 3x² – 5 | 3x⁴ + 6x³ – 2x² – 10x – 5 3x⁴ – 5x² ____________________ 6x³ + 3x² – 10x 6x³ – 10x ____________________ 3x² – 5 3x² – 5 ____________________ 0

Now factorize the quotient $x^2 + 2x + 1$:

$$x^2 + 2x + 1 = (x + 1)^2$$

Therefore, its zero is given by $x + 1 = 0$, i.e., $x = -1$ (repeated twice).

✔️ Other zeroes: $-1$ and $-1$

Q4

On dividing $x^3 – 3x^2 + x + 2$ by a polynomial $g(x)$, the quotient and remainder were $x – 2$ and $-2x + 4$, respectively. Find $g(x)$.

Solution

By Division Algorithm:

$$\text{Dividend} = \text{Divisor} \times \text{Quotient} + \text{Remainder}$$

Given:

• Dividend $p(x) = x^3 – 3x^2 + x + 2$

• Quotient $q(x) = x – 2$

• Remainder $r(x) = -2x + 4$

• Divisor $g(x) = ?$

$$x^3 – 3x^2 + x + 2 = g(x) \times (x – 2) + (-2x + 4)$$

$$x^3 – 3x^2 + x + 2 – (-2x + 4) = g(x) \times (x – 2)$$

$$x^3 – 3x^2 + x + 2 + 2x – 4 = g(x) \times (x – 2)$$

$$x^3 – 3x^2 + 3x – 2 = g(x) \times (x – 2)$$

Therefore, $g(x)$ is the quotient when we divide $x^3 – 3x^2 + 3x – 2$ by $(x – 2)$:

x² – x + 1 ____________________ x – 2 | x³ – 3x² + 3x – 2 x³ – 2x² ____________________ – x² + 3x – x² + 2x ____________________ x – 2 x – 2 ____________________ 0

✔️ $g(x) = x^2 – x + 1$

Q5

Give examples of polynomials $p(x)$, $g(x)$, $q(x)$ and $r(x)$, which satisfy the division algorithm and:

(i) deg $p(x)$ = deg $q(x)$

(ii) deg $q(x)$ = deg $r(x)$

(iii) deg $r(x)$ = 0

(i) deg $p(x)$ = deg $q(x)$

The degree of the quotient will be equal to the degree of the dividend only when the divisor is a constant.

Example: Divide $6x^2 + 2x + 2$ by 2.

• $p(x) = 6x^2 + 2x + 2$ (degree = 2)

• $g(x) = 2$ (constant)

• $q(x) = 3x^2 + x + 1$ (degree = 2)

• $r(x) = 0$

Verification: $p(x) = g(x) \times q(x) + r(x)$

$$6x^2 + 2x + 2 = 2(3x^2 + x + 1) + 0$$

✓ deg $p(x)$ = deg $q(x)$ = 2

(ii) deg $q(x)$ = deg $r(x)$

Example: Divide $x^3 + x$ by $x^2$.

• $p(x) = x^3 + x$

• $g(x) = x^2$

• $q(x) = x$ (degree = 1)

• $r(x) = x$ (degree = 1)

Verification: $p(x) = g(x) \times q(x) + r(x)$

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

✓ deg $q(x)$ = deg $r(x)$ = 1

(iii) deg $r(x)$ = 0

Degree of remainder is 0 when the remainder is a constant.

Example: Divide $x^3 + 1$ by $x^2$.

• $p(x) = x^3 + 1$

• $g(x) = x^2$

• $q(x) = x$

• $r(x) = 1$ (degree = 0)

Verification: $p(x) = g(x) \times q(x) + r(x)$

$$x^3 + 1 = (x^2 \times x) + 1 = x^3 + 1$$

✓ deg $r(x)$ = 0

NCERT Class 10 Maths