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

Question 1: In each of the following cases, find the remainder when

$\displaystyle \text{i) } x^4-3x^2+2x+1 \text{ is divided by } (x-1)$

$\displaystyle \text{ii) } x^3+3x^2-12x+4 \text{ is divided by } (x-2)$

$\displaystyle \text{iii) } x^4+1 \text{ is divided by } (x+1)$

Answer:

i) Required Remainder = Value of given polynomial $\displaystyle x^4-3x^2+2x+1 \text{ for } x = 1$

$\displaystyle \text{Therefore Remainder } = (1)^4-3(1)^2+2(1)+1 = 1$

ii) Required Remainder = Value of given polynomial $\displaystyle x^3+3x^2-12x+4 \text{ for } x = 2$

$\displaystyle \text{Therefore Remainder } = (2)^3+3(2)^2-12(2)+4 = 0$

iii) Required Remainder = Value of given polynomial $\displaystyle x^4+1 \text{ for } x = -1$

$\displaystyle \text{Therefore Remainder } = (-1)^4+1 = 2$

$\displaystyle \\$

Question 2: Show that

$\displaystyle \text{i) } (x-2) \text{ is a factor of } 5x^2+15x-50$

$\displaystyle \text{ii) } (3x+2) \text{ is a factor of } 3x^2-x-2$

Answer:

$\displaystyle \text{i) } \text{If } (x-2) \text{ is a factor of } 5x^2+15x-50$ , then the remainder should be $\displaystyle 0 \text{ for } x = 2$

$\displaystyle \text{Remainder } = 5(2)^2+15(2)-50 = 50-50=0$

$\displaystyle \text{Hence } (x-2) \text{ is a factor of } 5x^2+15x-50$

$\displaystyle \text{ii) } \text{If } (3x+2) \text{ is a factor of } 3x^2-x-2$ , then the remainder should be $\displaystyle 0 \text{ for } x = - \frac{2}{3}$

$\displaystyle \text{Remainder } = 3(- \frac{2}{3} )^2-(- \frac{2}{3} )-2 = \frac{4}{3} + \frac{2}{3} -2 = 0$

$\displaystyle \text{Hence } (3x+2) \text{ is a factor of } 3x^2-x-2$

$\displaystyle \\$

Question 3: Find which of the following is a factor of $\displaystyle 2x^3+3x^2-5x-6$

$\displaystyle \text{i) } (x+1)$ $\displaystyle \text{ii) } (2x-1)$ $\displaystyle \text{iii) } (x+2)$

Answer:

$\displaystyle \text{i) } \text{If } (x+1) \text{ is a factor of } 2x^3+3x^2-5x-6$ , then the remainder should be $\displaystyle 0 \text{ for } x =-1$

$\displaystyle \text{Remainder } = 2(-1)^3+3(-1)^2-5(-1)-6 = -2+3+5-6=0$

$\displaystyle \text{Hence } (x+1) \text{ is a factor of } 2x^3+3x^2-5x-6$

$\displaystyle \text{ii) } \text{If } (2x-1) \text{ is a factor of } 2x^3+3x^2-5x-6$ , then the remainder should be $\displaystyle 0 \text{ for } x = \frac{1}{2}$

$\displaystyle \text{Remainder } = 2( \frac{1}{2} )^3+3( \frac{1}{2} )^2-5( \frac{1}{2} )-6 = 1- \frac{3}{2} -6 \neq 0$

$\displaystyle \text{Hence } (2x-1)$ is NOT a factor of $\displaystyle 2x^3+3x^2-5x-6$

$\displaystyle \text{iii) } \text{If } (x+2) \text{ is a factor of } 2x^3+3x^2-5x-6$ , then the remainder should be $\displaystyle 0 \text{ for } x =-2$

$\displaystyle \text{Remainder } = 2(-2)^3+3(-2)^2-5(-2)-6 = -16+12+10-6= 0$

$\displaystyle \text{Hence } (x+2) \text{ is a factor of } 2x^3+3x^2-5x-6$

$\displaystyle \\$

Question 4: Find the value of $\displaystyle a \text{ or } k$ if

$\displaystyle \text{i) } (2x+1) \text{ is a factor of } 2x^2+ax-3$

$\displaystyle \text{ii) } (3x-4) \text{ is a factor of } 3x^2+2x-k$

$\displaystyle \text{iii) } (2x+1) \text{ is a factor of } (3k+2)x^3+(k-1)$

$\displaystyle \text{iv) } (x-2) \text{ is a factor of } 2x^5-6x^4-2ax^3+6ax^2+4ax+8$

Answer:

$\displaystyle \text{i) } (2x+1) \text{is a factor } \Rightarrow x = - \frac{1}{2}$

$\displaystyle \text{Therefore Remainder } =0 \text{ for } x = - \frac{1}{2}$

$\displaystyle \Rightarrow 2( - \frac{1}{2} )^2+a( - \frac{1}{2} )-3 = 0$

$\displaystyle \Rightarrow \frac{1}{2} - \frac{a}{2} -3=0$

$\displaystyle \Rightarrow 1-a-6=0$

$\displaystyle \Rightarrow a = 5$

$\displaystyle \text{ii) } (3x-4) \text{is a factor } \Rightarrow x = \frac{4}{3}$

$\displaystyle \text{Therefore Remainder } =0 \text{ for } x = \frac{4}{3}$

$\displaystyle \Rightarrow 3( \frac{4}{3} )^2+2( \frac{4}{3} )-k=0$

$\displaystyle \Rightarrow \frac{16}{3} + \frac{8}{3} -k=0$

$\displaystyle \Rightarrow k = 8$

$\displaystyle \text{iii) } \text{If } (2x+1) \text{is a factor } \Rightarrow x = - \frac{1}{2}$

$\displaystyle \text{Therefore Remainder } =0 \text{ for } x = - \frac{1}{2}$

$\displaystyle \Rightarrow (3k+2)(- \frac{1}{2} )^3+(k-1)=0$

$\displaystyle \Rightarrow -3k-2+8k-8=0$

$\displaystyle \Rightarrow k = 2$

$\displaystyle \text{iv) } \text{If } (x-2) \text{is a factor } \Rightarrow x = 2$

$\displaystyle \text{Therefore Remainder } = 0 \text{ for } x =2$

$\displaystyle \Rightarrow 2(2)^5-6(2)^4-2a(2)^3+6a(2)^2+4a(2)+8$

$\displaystyle \Rightarrow 64-96-16a+24a+8a+8 = 0$

$\displaystyle \Rightarrow a = \frac{24}{16} = \frac{3}{2}$

$\displaystyle \\$

Question 5: Find the value of $\displaystyle a \text{ and } b$ , when

$\displaystyle \text{i) } (x-2) \text{ and } (x+3) \text{ are both factors of } x^3+ax^2+bx-12$

$\displaystyle \text{ii) } (x-1) \text{ and } (x+2) \text{ are both factors of } x^3+(3a+1)x^2+bx-18$

Answer:

$\displaystyle \text{i) } \text{If } (x-2)$ is a factor

$\displaystyle \Rightarrow (2)^3+a(2)^2+b(2)-12=0$

$\displaystyle \Rightarrow 8+4a+2b-12=0$

$\displaystyle \Rightarrow 2a+b=2$ … … … … … i)

Similarly, $\displaystyle \text{If } (x+3)$ is a factor

$\displaystyle \Rightarrow (-3)^3+a(-3)^2+b(-3)-12=0$

$\displaystyle \Rightarrow -27+pa-3b-12=0$

$\displaystyle \Rightarrow 3a-b=13$ … … … … … ii)

Solving i) and ii) we get $\displaystyle a = 3 \text{ and } b =-4$

$\displaystyle \text{ii) } \text{If } (x-1)$ is a factor

$\displaystyle \Rightarrow (1)^3+(3a+1)(1)^2+b(1)-18=0$

$\displaystyle \Rightarrow 1 + 3a+1+b-18=0$

$\displaystyle \Rightarrow 3a+b=16$ … … … … … i)

$\displaystyle \text{If } (x+2)$ is a factor

$\displaystyle \Rightarrow (-2)^3+(3a+1)(-2)^2+b(-2)-18=k$

$\displaystyle \Rightarrow -8+12a+4-2b-18=0$

$\displaystyle \Rightarrow 6a-b=11$ … … … … … ii)

Solving i) and $\displaystyle \text{ii) } a = 3 \text{ and } b = 7$

$\displaystyle \\$

Question 6: $\displaystyle \text{When } x^3+2x^2-kx+4 \text{ is divided by } (x-2)$ , the remainder is $\displaystyle k$ . Find $\displaystyle k$ .

Answer:

$\displaystyle \text{When } x = 2,$ , $\displaystyle \text{Remainder } = 0$

$\displaystyle (2)^3+2(2)^2-k(2)+4 = k$

$\displaystyle \Rightarrow 20-2k = k$

$\displaystyle \Rightarrow k = \frac{20}{3}$

$\displaystyle \\$

Question 7: Find the value of $\displaystyle a$ , if the division of $\displaystyle ax^3+9x^2+4x-10$ by $\displaystyle (x+3)$ leaves a remainder of $\displaystyle 5$ .

Answer:

$\displaystyle \text{When } x = -3,$ , $\displaystyle \text{Remainder } = 5$

$\displaystyle a(-3)^3+9(-3)^2+4(-3)-10 = 5$

$\displaystyle \Rightarrow -27a+81-12-10=5$

$\displaystyle \Rightarrow 27a = 54$

$\displaystyle \Rightarrow a = 2$

$\displaystyle \\$

Question 8: $\displaystyle \text{If } x^3+ax^2+bx+6$ has $\displaystyle (x-2)$ as a factor and leaves a remainder of $\displaystyle 3$ when divided by $\displaystyle (x-3)$ , find the value of $\displaystyle a \text{ and } b$ . [2005]

Answer:

$\displaystyle \text{When } x=2$ , $\displaystyle \text{Remainder } = 0$

$\displaystyle \Rightarrow (2)^3+a(2)^2+b(2)+6 = 0$

$\displaystyle \Rightarrow 4a+2b = - 14$ … … … … … i)

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

$\displaystyle \Rightarrow (3)^3+a(3)^2+b(3)+6 = 3$

$\displaystyle \Rightarrow 9a+3b=-30$ … … … … … ii)

Solving i) and $\displaystyle \text{ii) } a = -3 \text{ and } b = -1$

$\displaystyle \\$

$\displaystyle \text{Question 9: Find the value of } a \text{ and } b \text{, When } 2x^3+ax^2+bx-2 \\ \\ \text{ leaves a Remainder } 7 \text{ and } 0 \text{ when divided by } (2x-3) \text{ and } (x+2) \text { respectively. }$

Answer:

$\displaystyle \text{When } x= \frac{3}{2} \text{, Remainder } = 7$

$\displaystyle \Rightarrow 2(\frac{3}{2})^3+a(\frac{3}{2})^2+b(\frac{3}{2})-2 = 7$

$\displaystyle \Rightarrow 27+9a+6b-8=28$

$\displaystyle \Rightarrow 3a+2b=3$ … … … … … i)

$\displaystyle \text{When } x = -2$ , $\displaystyle \text{Remainder } = 0$

$\displaystyle \Rightarrow 2(-2)^3+a(-2)^2+b(-2)-2= 0$

$\displaystyle \Rightarrow -16+4a-2b-2=0$

$\displaystyle \Rightarrow 2a-b=-9$ … … … … … ii)

Solving i) and $\displaystyle \text{ii) } a = 3 \text{ and } b = -3$

$\displaystyle \\$

Question 10: What number should be added to $\displaystyle 3x^3-5x^2+6x$ , so that when it is divided by $\displaystyle (x-3)$ , the remainder is $\displaystyle 8$ .

Answer:

$\displaystyle \text{Let } a$ be added to $\displaystyle 3x^3-5x^2+6x$ , so that when it is divided by $\displaystyle (x-3)$ , the remainder is $\displaystyle 8$

$\displaystyle \text{When } x =3$ , Remainder is $\displaystyle 8$

$\displaystyle \Rightarrow 3(3)^3-5(3)^2+6(3)+a=8$

$\displaystyle \Rightarrow 81-45+18+a=8$

$\displaystyle \Rightarrow a = -46$

$\displaystyle \\$

Question 11: What number should be subtracted to $\displaystyle x^3+3x^2-8x+14$ , so that when it is divided by $\displaystyle (x-2)$ , the remainder is $\displaystyle 10$ .

Answer:

$\displaystyle \text{Let } a$ be subtracted to $\displaystyle x^3+3x^2-8x+14$ , so that when it is divided by $\displaystyle (x-2)$ , the remainder is $\displaystyle 10$

$\displaystyle \text{When } x =2$ , Remainder is $\displaystyle 10$

$\displaystyle \Rightarrow (2)^3+3(2)^2-8(2)+14-a=10$

$\displaystyle \Rightarrow 8+12-16+14-a=10$

$\displaystyle \Rightarrow a = 8$

$\displaystyle \\$

Question 12: The polynomials $\displaystyle 2x^3-7x^2+ax-6$ and $\displaystyle x^3-8x^2+(2a+1)x-16$ leave the same remainder when divided by $\displaystyle (x-2)$ . Find the value of $\displaystyle a$ .

Answer:

$\displaystyle \text{For polynomial } 2x^3-7x^2+ax-6$ :

$\displaystyle \text{When } x =2 \text{, Remainder } = 2(2)^3-7(2)^2+a(2)-6 = 2a-18$

$\displaystyle \text{For polynomial } x^3-8x^2+(2a+1)x-16$ :

$\displaystyle \text{When } x =2 \text{, Remainder } = (2)^3-8(2)^2+(2a+1)(2)-16 = 4a-38$

Therefore $\displaystyle 2a-18=4a-38 \Rightarrow a = 10$

$\displaystyle \\$

Question 13: $\displaystyle \text{If } (x-2)$ is a factor of the expression $\displaystyle 2x^3+ax^2+bx-14$ and when the expression is divided by $\displaystyle (x-3)$ , it leaves a $\displaystyle \text{Remainder } 52$ . Find the value of $\displaystyle a \text{ and } b$ . [2013]