Class 10: Remainder and Factor Theorem – Exercise 11(b) – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Using the factor show that:

i) $(x-2)$ is a factor of $x^3-2x^2-9x+18$ . Hence factorise the polynomial $x^3-2x^2-9x+18$ .

ii) $(x+5)$ is a factor of $2x^3+5x^2-28x-15$ . Hence factorise the polynomial $2x^3+5x^2-28x-15$ .

iii) $(3x+2)$ is a factor of $3x^3+2x^2-3x-2$ . Hence factorise the polynomial $3x^3+2x^2-3x-2$ .

iv) $(2x+7)$ is a factor of $2x^3+5x^2-11x-14$ . Hence factorise the polynomial $2x^3+5x^2-11x-14$ .

Answer:

i) Since $x -2 = 0 \Rightarrow x = 2$

Remainder $= (2)^3-2(2)^2-9(2)+18 = 8-8-18+18=0$

Therefore $\Rightarrow$ $(x-2)$ is a factor of $x^3-2x^2-9x+18$ .

Dividing $x^3-2x^2-9x+18$ by $(x-2)$

$x-2 ) \overline {x^3-2x^2-9x+18} (x^2-9$
$(-) \ \ \underline {x^3-2x^2}$
$-9x+18$
$(-) \ \ \underline{-9x+18 }$
$\times$

Therefore $x^3-2x^2-9x+18 = (x-2)(x^2-9)=(x-2)(x-3)(x+3)$

$\\$

ii) Since $x +5 = 0 \Rightarrow x = -5$

Remainder $=2(-5)^3+5(-5)^2-28(-5)-15 = -250+125+140-15=0$

Therefore $\Rightarrow$ $(x+5)$ is a factor of $2x^3+5x^2-28x-15$ .

Dividing $2x^3+5x^2-28x-15$ by $(x+5)$

$x+5 ) \overline {2x^3+5x^2-28x-15} (2x^2-5x-3$
$(-) \ \ \underline {2x^3+10x^2}$
$-5x^2-28x-15$
$(-) \ \ \underline{-5x^2-25x}$
$-3x-15$
$(-) \ \ \underline{ -3x-15}$
$\times$

$2x^3+5x^2-28x-15$

$= (x+5)(2x^2-5x-3)$

$= (x+5)(2x^2-6x+x-3)$

$=(x+5)[2x(x-3)+(x-3)]$

$= (x+5)(x-3)(2x+1)$

Hence $2x^3+5x^2-28x-15 =(x+5)(x-3)(2x+1)$

$\\$

iii) Since $x =-\frac{2}{3}$

Remainder $= 3(-\frac{2}{3})^3+2(-\frac{2}{3})^2-3(-\frac{2}{3})-2 = -\frac{8}{9}+\frac{8}{9}+2-2 = 0$

Therefore $(3x+2)$ is a factor of $3x^3+2x^2-3x-2$

Dividing $3x^3+2x^2-3x-2$ by $(3x+2)$

$3x+2 ) \overline {3x^3+2x^2-3x-2} (x^2-1$
$(-) \ \ \underline {3x^3+2x^2}$
$-3x-2$
$(-) \ \ \underline{-3x-2 }$
$\times$

$3x^3+2x^2-3x-2 = (3x+2)(x^2-1) = (3x+2)(x-1)(x+1)$

Hence $3x^3+2x^2-3x-2= (3x+2)(x-1)(x+1)$

$\\$

iv) Since $x = -\frac{7}{2}$

Remainder $= 2(-\frac{7}{2})^3+5(-\frac{7}{2})^2-11(-\frac{7}{2})-14$

$= -\frac{343}{4}+\frac{245}{4}+\frac{77}{2}-14= \frac{-343+245+154-56}{4}=0$

Therefore $(2x+7)$ is a factor of $2x^3+5x^2-11x-14$

Dividing $2x^3+5x^2-11x-14$ by $(2x+7)$

$2x+7 ) \overline {2x^3+5x^2-11x-14} (x^2-x-2$
$(-) \ \ \underline {2x^3+7x^2}$
$-2x^2-11x-14$
$(-) \ \ \underline{-2x^2-7x}$
$-4x-14$
$(-) \ \ \underline{ -2x-14}$
$\times$

$2x^3+5x^2-11x-14 = (2x+7)(x^2-x-2)$

$= (2x+7)(x^2-2x+x-2)$

$=(2x+7)[x(x-2)+(x-2)]$

$= (2x+7)(x-2)(x+1)$

Hence $2x^3+5x^2-11x-14 = (2x+7)(x-2)(x+1)$

$\\$

Question 2: Factorise using factor theorem:

i) $3x^3+2x^2-19x+6$ [2012]

ii) $2x^3+x^2-13x+6$

iii) $3x^3+2x^2-23x-30$

iv) $4x^3+7x^2-36x-63$

v) $x^3+x^2-4x-4$ [2004]

vi) $2x^3-7x^2-3x+18$

vii) $3x^3+10x^2+x-6$

Answer:

i) $3x^3+2x^2-19x+6$

For $x = 2$ ,

Expression: $= 3(2)^3+2(2)^2-19(2)+6 = 24+8-38+6=0$

Hence $(x-2)$ is a factor of $3x^3+2x^2-19x+6$

$x-2 ) \overline {3x^3+2x^2-19x+6} (3x^2+8x-3$
$(-) \ \ \underline {3x^3-6x^2}$
$8x^2-19x+6$
$(-) \ \ \underline{8x^2-16x}$
$-3x+6$
$(-) \ \ \underline{ -3x+6}$
$\times$

$3x^3+2x^2-19x+6 = (x-2)(3x^2+8x-3)$

$= (x-2)(3x^2+9x-x-3)$

$=(x-2)[3x(x+3)-(x+3)]$

$= (x-2)(x+3)(3x-1)$

Hence $3x^3+2x^2-19x+6= (x-2)(x+3)(3x-1)$

$\\$

ii) $2x^3+x^2-13x+6$

For $x = 2$ ,

Expression: $= 2(2)^3+(2)^2-13(2)+6 = 16+4-26+6 = 0$

Hence $(x-2)$ is a factor of $2x^3+x^2-13x+6$

$x-2 ) \overline {2x^3+x^2-13x+6} (2x^2+5x-3$
$(-) \ \ \underline {2x^3-4x^2}$
$5x^2-13x+6$
$(-) \ \ \underline{5x^2-10x}$
$-3x+6$
$(-) \ \ \underline{ -3x+6}$
$\times$

$2x^3+x^2-13x+6 = (x-2)(2x^2+6x-x-3)$

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

$= (x-2)[2x(x+3)-(x+3)]$

$=(x-2)(x+3)(2x-1)$

Hence $2x^3+x^2-13x+6=(x-2)(x+3)(2x-1)$

$\\$

iii) $3x^3+2x^2-23x-30$

For $x = -2$ ,

Expression: $3(-2)^3+2(-2)^2-23(-2)-30 = -24+8+46-30=0$

Therefore $(x+2)$ is a factor of $3x^3+2x^2-23x-30$

$x+2 ) \overline {3x^3+2x^2-23x-30} (3x^2-4x-15$
$(-) \ \ \underline {3x^3+6x^2}$
$-4x^2-23x-30$
$(-) \ \ \underline{-4x^2-8x}$
$-15x-30$
$(-) \ \ \underline{ -15x-30}$
$\times$

$x^3+2x^2-23x-30 = (x+2)(3x^2-4x-15)$

$= (x+2)(3x^2-9x+5x-15)$

$= (x+2)[3x(x-3)+5(x-3)]$

$=(x+2)(3x+5)(x-3)$

Hence $3x^3+2x^2-23x-30=(x+2)(3x+5)(x-3)$

$\\$

iv) $4x^3+7x^2-36x-63$

For $x = -3$ ,

Expression: $4(-3)^3+7(-3)^2-36(-3)-63 = -108+63+108-63 = 0$

Therefore $(x+3)$ is a factor of $4x^3+7x^2-36x-63$

$x+3 ) \overline {4x^3+7x^2-36x-63} (4x^2-5x-21$
$(-) \ \ \underline {4x^3+12x^2}$
$-5x^2-36x-63$
$(-) \ \ \underline{-5x^2-15x}$
$-21x-63$
$(-) \ \ \underline{ -21x-63}$
$\times$

$4x^3+7x^2-36x-63 = (x+3)(4x^2-5x-21)$

$= (x+3)(4x^2-5x-21)$

$= (x+3)[4x(x-3)+7(x-3)]$

$= (x+3)(x-3)(4x+7)$

Hence $4x^3+7x^2-36x-63 = (x+3)(x-3)(4x+7)$

$\\$

v) $x^3+x^2-4x-4$

For $x = -1$

Expression: $(-1)^3+(-1)^2-4(-1)-4 = -1+1+4-4 = 0$

Therefore $(x+1)$ is a factor of $x^3+x^2-4x-4$

$x+1 ) \overline {x^3+x^2-4x-4} (x^2-4$
$(-) \ \ \underline {x^3+x^2}$
$-4x-4$
$(-) \ \ \underline{-4x-4 }$
$\times$

$x^3+x^2-4x-4 = (x+1)(x^2-4)$

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

Hence $x^3+x^2-4x-4= (x+1)(x-2)(x+2)$

$\\$

vi) $2x^3-7x^2-3x+18$

For $x = 2$

Expression: $2(2)^3-7(2)^2-3(2)+18 = 16-28-6+18= 0$

Therefore $(x-2)$ is a factor of $2x^3-7x^2-3x+18$

$x-2 ) \overline {2x^3-7x^2-3x+18} (2x^2-3x-9$
$(-) \ \ \underline {2x^3-4x^2}$
$-3x^2-3x+18$
$(-) \ \ \underline{-3x^2+6x}$
$-9x+18$
$(-) \ \ \underline{ -9x+18}$
$\times$

$2x^3-7x^2-3x+18 = (x-2)(2x^2-3x-9)$

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

$= (x-2)[2x(x-3)+3(x-3)]$

$= (x-2)(x-3)(2x+3)$

Hence $2x^3-7x^2-3x+18= (x-2)(x-3)(2x+3)$

$\\$

vii) $3x^3+10x^2+x-6$

For $x = -1$

Expression: $3(-1)^3+10(-1)^2+(-1)-6= -31=10-1-6= 0$

Therefore $(x+1)$ is a factor of $3x^3+10x^2+x-6$

$x+1 ) \overline {3x^3+10x^2+x-6} (3x^2+7x-6$
$(-) \ \ \underline {3x^3+3x^2}$
$7x^2+x-6$
$(-) \ \ \underline{7x^2+7x}$
$-6x-6$
$(-) \ \ \underline{ -6x-6}$
$\times$

$3x^3+10x^2+x-6 = (x+1)(3x^2+7x-6)$

$= (x+1)(3x^2+9x-2x-6)$

$= (x+1)[3x(3x+3)-2(x+3)]$

$= (x+1)(3x+3)(3x-2)$

Hence $3x^3+10x^2+x-6= (x+1)(3x+3)(3x-2)$

$\\$

Question 3: If $(x-2)$ and $(x+1)$ are factors of $x^3+3x^2+ax+b$ , calculate the values of $a \ and \ b$ . Then factorise.

Answer:

When $x = 2$

$f(x)=(2)^3+3(2)^2+a(2)+b \Rightarrow = 2a+b=-20$ … … … … … i)

When $x = -1$

$f(x) =(-1)^3+3(-1)^2+a(-1)+b \Rightarrow a-b=2$ … … … … … ii)

Solving i) and ii) we get $a = -6 \ and \ b = -8$

$\\$

Question 4: When $4x^3-bx^2+x-c$ is divided by $(x+1)$ and $(2x-3)$ the remainder left is $0 \ and \ 30$ respectively. Calculate the values of $a \ and \ b$ and then factorise the given polynomial.

Answer:

When $x =-1$ , Remainder $= 0$

$\displaystyle \text{Therefore } 4(-1)^3-b(-1)^2+(-1)-c=0 \Rightarrow b+c=-5$ … … … … … i)

$\displaystyle \text{When } x = \frac{3}{2} , Remainder = 30$

$\displaystyle \text{Therefore } 4(\frac{3}{2})^3-b(\frac{3}{2})^2+(\frac{3}{2})-c=0 \Rightarrow 9b+4c=-60$ … … … … … i)

Solving i) and ii) we get $b = -8 \ and \ b = 3$

$\\$

Question 5: If $(x+a)$ is a common factor of expressions $f(x)=x^2+px+q$ and $g(x)=x^2+mx+n$ ; show that $\displaystyle a = \frac{n-q}{m-p}$

Answer:

When $x =-a$ , Remainder $= 0$

$\therefore f(x)=(-a)^2+p(-a)+q=0 \Rightarrow a^2-pa+q= 0 \Rightarrow a^2=pa-q$ … … … i)

$\therefore x(x)=(-a)^2+m(-a)+n=0 \Rightarrow a^2-ma+n= 0 \Rightarrow a^2=ma-n$ … … … ii)

From i) and ii)

$\displaystyle pa-q=ma-n \Rightarrow a = \frac{n-q}{m-p}$

$\\$

Question 6: When $ax^3+3x^2-3$ and $2x^3-5x+a$ are divided by $(x-4)$ , the remainder is the same. Find the value of $a$ .

Answer:

When $x = 4$

$\text{Remainder}_1 = a(4)^3+3(4)^2-3 = 64a+45$

$\text{Remainder}_2 = 2(4)^3-5(4)+a = 108+a$

$\text{Now Remainder}_1 = \text{Remainder}_2$

$64a+45=108+a \Rightarrow a = 1$

$\\$

Question 7: Find the value of $a$ , if $(x-a)$ is a factor of $x^3-ax^2+x+2$ . [2003]

Answer:

When $x = a$ , Remainder $= 0$

Therefore $(a)^3-a(a)^2+(a)+2 =0$

$\Rightarrow a^3-a^3+a+2=0$

$\Rightarrow a = 2$

$\\$

Question 8: What number should be subtracted from $3x^3+x^2-22x+15$ , so that $(x+3)$ is a factor of the given polynomial.

Answer:

Let $a$ be subtracted from the polynomial.

When $x = -3$ , Remainder $= 0$

Therefore $3x^3+x^2-22x+15-a = 0$

$\Rightarrow 3(-3)^3+(-3)^2-22(-3)+15-a = 0$

$\Rightarrow -81+90-a=0$

$\Rightarrow a = 9$

$\\$

ICSE / ISC / CBSE Mathematics Portal for K12 Students

Mathematics Made Easy for School Students

Class 10: Remainder and Factor Theorem – Exercise 11(b)

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