Polynomials
NCERT Solutions • Class 9 Maths • Chapter 2Exercise 2.3
1. Determine which of the following polynomials has $(x + 1)$ a factor:
(i) $x^3 + x^2 + x + 1$
Let $p(x) = x^3 + x^2 + x + 1$. Since divisor is $x+1$, put $x = -1$.
$p(-1) = (-1)^3 + (-1)^2 + (-1) + 1 = -1 + 1 – 1 + 1 = 0$
Yes, it is a factor
Let $p(x) = x^3 + x^2 + x + 1$. Since divisor is $x+1$, put $x = -1$.
$p(-1) = (-1)^3 + (-1)^2 + (-1) + 1 = -1 + 1 – 1 + 1 = 0$
Yes, it is a factor
(ii) $x^4 + x^3 + x^2 + x + 1$
$p(-1) = (-1)^4 + (-1)^3 + (-1)^2 + (-1) + 1 = 1 – 1 + 1 – 1 + 1 = 1 \neq 0$
No, it is not a factor
$p(-1) = (-1)^4 + (-1)^3 + (-1)^2 + (-1) + 1 = 1 – 1 + 1 – 1 + 1 = 1 \neq 0$
No, it is not a factor
(iii) $x^4 + 3x^3 + 3x^2 + x + 1$
$p(-1) = (-1)^4 + 3(-1)^3 + 3(-1)^2 + (-1) + 1 = 1 – 3 + 3 – 1 + 1 = 1 \neq 0$
No, it is not a factor
$p(-1) = (-1)^4 + 3(-1)^3 + 3(-1)^2 + (-1) + 1 = 1 – 3 + 3 – 1 + 1 = 1 \neq 0$
No, it is not a factor
(iv) $x^3 – x^2 – (2 + \sqrt{2})x + \sqrt{2}$
$p(-1) = (-1)^3 – (-1)^2 – (2+\sqrt{2})(-1) + \sqrt{2} = -1 – 1 + 2 + \sqrt{2} + \sqrt{2} = 2\sqrt{2} \neq 0$
No, it is not a factor
$p(-1) = (-1)^3 – (-1)^2 – (2+\sqrt{2})(-1) + \sqrt{2} = -1 – 1 + 2 + \sqrt{2} + \sqrt{2} = 2\sqrt{2} \neq 0$
No, it is not a factor
2. Use the Factor Theorem to determine whether $g(x)$ is a factor of $p(x)$ in each case:
(i) $p(x) = 2x^3 + x^2 – 2x – 1, \quad g(x) = x + 1$
Zero of $g(x)$ is $-1$. Check $p(-1)$.
$p(-1) = 2(-1)^3 + (-1)^2 – 2(-1) – 1 = -2 + 1 + 2 – 1 = 0$
Yes, g(x) is a factor
Zero of $g(x)$ is $-1$. Check $p(-1)$.
$p(-1) = 2(-1)^3 + (-1)^2 – 2(-1) – 1 = -2 + 1 + 2 – 1 = 0$
Yes, g(x) is a factor
(ii) $p(x) = x^3 + 3x^2 + 3x + 1, \quad g(x) = x + 2$
Zero of $g(x)$ is $-2$. Check $p(-2)$.
$p(-2) = (-2)^3 + 3(-2)^2 + 3(-2) + 1 = -8 + 12 – 6 + 1 = -1 \neq 0$
No, g(x) is not a factor
Zero of $g(x)$ is $-2$. Check $p(-2)$.
$p(-2) = (-2)^3 + 3(-2)^2 + 3(-2) + 1 = -8 + 12 – 6 + 1 = -1 \neq 0$
No, g(x) is not a factor
(iii) $p(x) = x^3 – 4x^2 + x + 6, \quad g(x) = x – 3$
Zero of $g(x)$ is $3$. Check $p(3)$.
$p(3) = (3)^3 – 4(3)^2 + 3 + 6 = 27 – 36 + 3 + 6 = 0$
Yes, g(x) is a factor
Zero of $g(x)$ is $3$. Check $p(3)$.
$p(3) = (3)^3 – 4(3)^2 + 3 + 6 = 27 – 36 + 3 + 6 = 0$
Yes, g(x) is a factor
3. Find the value of $k$, if $x – 1$ is a factor of $p(x)$ in each of the following cases:
(i) $p(x) = x^2 + x + k$
Since $x-1$ is a factor, $p(1) = 0$.
$(1)^2 + 1 + k = 0 \Rightarrow 1 + 1 + k = 0 \Rightarrow$ $k = -2$
Since $x-1$ is a factor, $p(1) = 0$.
$(1)^2 + 1 + k = 0 \Rightarrow 1 + 1 + k = 0 \Rightarrow$ $k = -2$
(ii) $p(x) = 2x^2 + kx + \sqrt{2}$
$p(1) = 0 \Rightarrow 2(1)^2 + k(1) + \sqrt{2} = 0 \Rightarrow 2 + k + \sqrt{2} = 0$
$k = -(2 + \sqrt{2})$
$p(1) = 0 \Rightarrow 2(1)^2 + k(1) + \sqrt{2} = 0 \Rightarrow 2 + k + \sqrt{2} = 0$
$k = -(2 + \sqrt{2})$
(iii) $p(x) = kx^2 – \sqrt{2}x + 1$
$p(1) = 0 \Rightarrow k(1)^2 – \sqrt{2}(1) + 1 = 0 \Rightarrow k – \sqrt{2} + 1 = 0$
$k = \sqrt{2} – 1$
$p(1) = 0 \Rightarrow k(1)^2 – \sqrt{2}(1) + 1 = 0 \Rightarrow k – \sqrt{2} + 1 = 0$
$k = \sqrt{2} – 1$
(iv) $p(x) = kx^2 – 3x + k$
$p(1) = 0 \Rightarrow k(1)^2 – 3(1) + k = 0 \Rightarrow k – 3 + k = 0 \Rightarrow 2k = 3$
$k = \frac{3}{2}$
$p(1) = 0 \Rightarrow k(1)^2 – 3(1) + k = 0 \Rightarrow k – 3 + k = 0 \Rightarrow 2k = 3$
$k = \frac{3}{2}$
4. Factorise:
(i) $12x^2 – 7x + 1$
Split $-7x$ into $-4x – 3x$.
$= 12x^2 – 4x – 3x + 1 = 4x(3x – 1) – 1(3x – 1) =$ $(3x – 1)(4x – 1)$
Split $-7x$ into $-4x – 3x$.
$= 12x^2 – 4x – 3x + 1 = 4x(3x – 1) – 1(3x – 1) =$ $(3x – 1)(4x – 1)$
(ii) $2x^2 + 7x + 3$
Split $7x$ into $6x + 1x$.
$= 2x^2 + 6x + x + 3 = 2x(x + 3) + 1(x + 3) =$ $(x + 3)(2x + 1)$
Split $7x$ into $6x + 1x$.
$= 2x^2 + 6x + x + 3 = 2x(x + 3) + 1(x + 3) =$ $(x + 3)(2x + 1)$
(iii) $6x^2 + 5x – 6$
Split $5x$ into $9x – 4x$.
$= 6x^2 + 9x – 4x – 6 = 3x(2x + 3) – 2(2x + 3) =$ $(2x + 3)(3x – 2)$
Split $5x$ into $9x – 4x$.
$= 6x^2 + 9x – 4x – 6 = 3x(2x + 3) – 2(2x + 3) =$ $(2x + 3)(3x – 2)$
(iv) $3x^2 – x – 4$
Split $-x$ into $-4x + 3x$.
$= 3x^2 + 3x – 4x – 4 = 3x(x + 1) – 4(x + 1) =$ $(x + 1)(3x – 4)$
Split $-x$ into $-4x + 3x$.
$= 3x^2 + 3x – 4x – 4 = 3x(x + 1) – 4(x + 1) =$ $(x + 1)(3x – 4)$
5. Factorise:
(i) $x^3 – 2x^2 – x + 2$
$= x^2(x – 2) – 1(x – 2) = (x^2 – 1)(x – 2)$
$= (x – 1)(x + 1)(x – 2)$ $\rightarrow$ $(x-1)(x+1)(x-2)$
$= x^2(x – 2) – 1(x – 2) = (x^2 – 1)(x – 2)$
$= (x – 1)(x + 1)(x – 2)$ $\rightarrow$ $(x-1)(x+1)(x-2)$
(ii) $x^3 – 3x^2 – 9x – 5$
Factor is $x+1$. Divide polynomial by $(x+1)$ to get $(x^2 – 4x – 5)$.
$x^2 – 4x – 5 = (x-5)(x+1)$. $\rightarrow$ $(x+1)(x+1)(x-5)$
Factor is $x+1$. Divide polynomial by $(x+1)$ to get $(x^2 – 4x – 5)$.
$x^2 – 4x – 5 = (x-5)(x+1)$. $\rightarrow$ $(x+1)(x+1)(x-5)$
(iii) $x^3 + 13x^2 + 32x + 20$
Factor is $x+1$. Divide by $(x+1)$ to get $(x^2 + 12x + 20)$.
$x^2 + 12x + 20 = (x+10)(x+2)$. $\rightarrow$ $(x+1)(x+2)(x+10)$
Factor is $x+1$. Divide by $(x+1)$ to get $(x^2 + 12x + 20)$.
$x^2 + 12x + 20 = (x+10)(x+2)$. $\rightarrow$ $(x+1)(x+2)(x+10)$
(iv) $2y^3 + y^2 – 2y – 1$
$= y^2(2y + 1) – 1(2y + 1) = (y^2 – 1)(2y + 1)$
$= (y – 1)(y + 1)(2y + 1)$ $\rightarrow$ $(y-1)(y+1)(2y+1)$
$= y^2(2y + 1) – 1(2y + 1) = (y^2 – 1)(2y + 1)$
$= (y – 1)(y + 1)(2y + 1)$ $\rightarrow$ $(y-1)(y+1)(2y+1)$