Exercise-2.3, Class 9th, Maths, Chapter 2, NCERT

1. Determine which of the following polynomials has $(x+1)$ as a factor:

(i) $x^3+x^2+x+1$
Check $p(-1)$
$p(-1)=(-1{)}^3+(-1{)}^2+(-1)+1=-1+1-1+1=0$
So $(x+1)$ is a factor.
Factorisation: divide or compare coefficients:

$$x^3+x^2+x+1=(x+1)(x^2+1)\mathrm{}$$

(ii) $x^4+x^3+x^2+x+1$
$p(-1)=1-1+1-1+1=1\mathrm{\ne }0$
So $(x+1)$ is not a factor.

(iii) $x^4+3x^3+3x^2+x+1$
$p(-1)=1-3+3-1+1=1\mathrm{\ne }0$
So $(x+1)$ is not a factor.

(iv) $x^3-x^2-(2+\sqrt{2})x+\sqrt{2}$
$p(-1)=(-1{)}^3-(-1{)}^2-(2+\sqrt{2})(-1)+\sqrt{2}$

$=-1-1+(2+\sqrt{2})+\sqrt{2}=2\sqrt{2}\mathrm{\ne }0$So $(x+1)$ is not a factor.

Answer (Q1): Only (i) has $(x+1)$ as a factor.

iemh102

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2. Use the Factor Theorem to determine whether $g(x)$ is a factor of $p(x)$ in each case:

(i) $p(x)=2x^3+x^2-2x-1,g(x)=x+1$.
Root of $g(x)$ is $x=-1$. Evaluate $p(-1)$:
$p(-1)=2(-1{)}^3+(-1{)}^2-2(-1)-1=-2+1+2-1=0$
So $x+1$ is a factor. (Yes.)

(ii) $p(x)=x^3+3x^2+3x+1,g(x)=x+2$
Root is $x=-2$. Evaluate $p(-2)=(-8)+12-6+1=-1\mathrm{\ne }0$
So $x+2$ is not a factor. (No.)

(iii) $p(x)=x^3-4x^2+x+6,g(x)=x-3$
Root is $x=3$. Evaluate $p(3)=27-36+3+6=0$
So $x-3$ is a factor. (Yes.)

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3. Find the value of $k$ if $x-1$ is a factor of $p(x)$ in each case (so set $p(1)=0$):

(i) $p(x)=x^2+x+k$
$p(1)=1+1+k=0\Rightarrow k=-2.$

(ii) $p(x)=2x^2+kx+2$
$p(1)=2+k+2=0\Rightarrow k=-4$

(iii) $p(x)=k{x}^2-2x+1$
$p(1)=k-2+1=0\Rightarrow k=1$

(iv) $p(x)=k{x}^2-3x+k$
$p(1)=k-3+k=0\Rightarrow 2k-3=0\Rightarrow k=\frac{3}{2}\mathrm{}$

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4. Factorise the following quadratics:

(i) $12x^2-7x+1$
Find pair for $12\cdot 1=12$ that sum to $-7$: $-3$ and $-4$
Split middle term: $12x^2-3x-4x+1=3x(4x-1)-1(4x-1)=(4x-1)(3x-1)$

(ii) $2x^2+7x+3$
Product $2\cdot 3=6$, sum $7$: $6$ and $1$.
Split: $2x^2+6x+x+3=2x(x+3)+1(x+3)=(2x+1)(x+3)$

(iii) $6x^2+5x-6$
Product $6\cdot (-6)=-36$, sum $5$: $9$ and $-4$.
Split: $6x^2+9x-4x-6=3x(2x+3)-2(2x+3)=(2x+3)(3x-2)$

(iv) $3x^2-x-4$
Product $3\cdot (-4)=-12$, sum $-1$: $-4$ and $3$.
Split: $3x^2-4x+3x-4=x(3x-4)+1(3x-4)=(3x-4)(x+1)$

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5. Factorise the following cubics / polynomials:

(i) $x^3-2x^2-x+2$
Group: $x^2(x-2)-1(x-2)=(x-2)(x^2-1)=(x-2)(x-1)(x+1)$

(ii) $x^3-3x^2-9x-5$
Try rational roots ±1,±5. $p(5)=125-75-45-5=0$ so $x=5$ is a root.
Divide by $(x-5)$ → quotient $x^2+2x+1=(x+1{)}^2$
So factorisation: $(x-5)(x+1{)}^2$

(iii) $x^3+13x^2+32x+20$
Test $x=-1$: $-1+13-32+20=0$ → factor $(x+1)$. Divide gives $x^2+12x+20$
Quadratic factors: $x^2+12x+20=(x+10)(x+2)$
So overall: $(x+1)(x+10)(x+2)$

(iv) $2y^3+y^2-2y-1$
Group: $y^2(2y+1)-1(2y+1)=(2y+1)(y^2-1)=(2y+1)(y-1)(y+1)$