Class 10 – Factorization – ICSE Board Problems (1)

$\displaystyle \textbf{Question 1: }\text{Find the remainder, when }2x^3-3x^2+7x-8\text{ is divided by } \\ (x-1).\hfill \text{[ICSE 2000]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=2x^3-3x^2+7x-8$
$\displaystyle \text{Since }f(x)\text{ is divided by }(x-1),\text{ by the Remainder Theorem, the remainder is }f(1).$
$\displaystyle f(1)=2(1)^3-3(1)^2+7(1)-8$
$\displaystyle =2-3+7-8=-2$
$\displaystyle \therefore \text{The required remainder is }-2.$
$\\$

$\displaystyle \textbf{Question 2: }\text{A function }f\text{ is defined by }f(x)=144-16x^2,\text{ calculate }f(2).$
$\displaystyle \text{Also, find the value of }x,\text{ when }f(x)=0.\hfill \text{[ICSE 2001]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }f(x)=144-16x^2$
$\displaystyle \text{Putting }x=2,\text{ we get}$
$\displaystyle f(2)=144-16(2)^2$
$\displaystyle =144-16\times4$
$\displaystyle =144-64=80$
$\displaystyle \text{Also, given }f(x)=0$
$\displaystyle 144-16x^2=0$
$\displaystyle 16x^2=144$
$\displaystyle x^2=\frac{144}{16}=9$
$\displaystyle x=\pm3$
$\displaystyle \therefore f(2)=80\text{ and }x=\pm3.$
$\\$

$\displaystyle \textbf{Question 3: }\text{Find the value of }a,\text{ if }(x-a)\text{ is a factor of }x^3-ax^2+x+2.\hfill \text{[ICSE 2003]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=x^3-ax^2+x+2$
$\displaystyle \text{Since }(x-a)\text{ is a factor of }f(x),\text{ by the Factor Theorem, }f(a)=0.$
$\displaystyle f(a)=a^3-a(a)^2+a+2=0$
$\displaystyle a^3-a^3+a+2=0$
$\displaystyle a+2=0$
$\displaystyle a=-2$
$\displaystyle \therefore \text{The value of }a\text{ is }-2.$
$\\$

$\displaystyle \textbf{Question 4: }\text{If }x^3+ax^2+bx+6\text{ has }(x-2)\text{ as a factor and leaves a}$
$\displaystyle \text{remainder of }3\text{ when divided by }(x-3),\text{ find the value of }a\text{ and }b.\hfill \text{[ICSE 2005]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=x^3+ax^2+bx+6$
$\displaystyle \text{Since }(x-2)\text{ is a factor of }f(x),\text{ by the Factor Theorem, }f(2)=0.$
$\displaystyle 2^3+a(2)^2+b(2)+6=0$
$\displaystyle 8+4a+2b+6=0$
$\displaystyle 4a+2b=-14$
$\displaystyle 2a+b=-7\qquad \cdots (i)$
$\displaystyle \text{Also, the remainder when }f(x)\text{ is divided by }(x-3)\text{ is }3.$
$\displaystyle \therefore f(3)=3$
$\displaystyle 3^3+a(3)^2+b(3)+6=3$
$\displaystyle 27+9a+3b+6=3$
$\displaystyle 9a+3b=-30$
$\displaystyle 3a+b=-10\qquad \cdots (ii)$
$\displaystyle \text{Subtracting }(i)\text{ from }(ii),$
$\displaystyle a=-3$
$\displaystyle \text{Substituting }a=-3\text{ in }(i),$
$\displaystyle 2(-3)+b=-7$
$\displaystyle -6+b=-7$
$\displaystyle b=-1$
$\displaystyle \therefore a=-3\text{ and }b=-1.$
$\\$

$\displaystyle \textbf{Question 5: }\text{When divided by }(x-3)\text{ the polynomials }x^3-px^2+x+6\text{ and }$
$\displaystyle 2x^3-x^2-(p+3)x-6\text{ leave the same remainder. Find the value of }p.\hfill \text{[ICSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=x^3-px^2+x+6\text{ and }g(x)=2x^3-x^2-(p+3)x-6$
$\displaystyle \text{By the Remainder Theorem, the remainder on dividing by }(x-3)\text{ is obtained by putting }x=3.$
$\displaystyle \text{Remainder}_1=f(3)=3^3-p(3)^2+3+6$
$\displaystyle =27-9p+3+6=36-9p$
$\displaystyle \text{Remainder}_2=g(3)=2(3)^3-(3)^2-(p+3)(3)-6$
$\displaystyle =54-9-3p-9-6=30-3p$
$\displaystyle \text{Since the remainders are equal,}$
$\displaystyle 36-9p=30-3p$
$\displaystyle 6=6p$
$\displaystyle p=1$
$\displaystyle \therefore \text{The value of }p\text{ is }1.$
$\\$

Question 6: Use the remainder theorem to factorize the following expression: $2x^3+x^2-13x+6$ . [ICSE2010]
$\displaystyle \text{Answer:}$
Let $x =2$
Remainder $= 2(2)^3+(2)^2-13(2)+6 = 16+4-26+6=0$
Hence $(x-2)$ is a factor of $2x^3+x^2-13x+6$

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

$2x^3+x^2-13x+6 = (x-2)(2x^2+5x-3)$
$= (x-2)(2x^2+6x-x-3)$
$= (x-2)[2x(x+3)-(x+3)]$
$= (x-2)(x+3)(2x-1)$
Hence $2x^3+x^2-13x+6 = (x-2)(x+3)(2x-1)$

$\displaystyle \textbf{Question 7: }\text{When divided by }(x-3),\text{ the polynomials}$
$\displaystyle x^3-px^2+x+6\text{ and }2x^3-x^2-(p+3)x-6\text{ leave the same remainder.}$
$\displaystyle \text{Find the value of }p.\hfill \text{[ICSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=x^3-px^2+x+6$
$\displaystyle \text{and }g(x)=2x^3-x^2-(p+3)x-6$
$\displaystyle \text{Since the given polynomials leave the same remainder when divided by }(x-3),$
$\displaystyle f(3)=g(3)$
$\displaystyle 3^3-p(3)^2+3+6=2(3)^3-(3)^2-(p+3)(3)-6$
$\displaystyle 27-9p+3+6=54-9-3(p+3)-6$
$\displaystyle 36-9p=39-3(p+3)$
$\displaystyle 36-9p=39-3p-9$
$\displaystyle 36-9p=30-3p$
$\displaystyle 36-30=-3p+9p$
$\displaystyle 6=6p$
$\displaystyle p=1$
$\displaystyle \therefore \text{The value of }p\text{ is }1.$
$\\$

$\displaystyle \textbf{Question 8: }\text{Find the value of }k\text{ if }(x-2)\text{ is a factor of }x^3+2x^2-kx+10.$
$\displaystyle \text{Hence determine whether }(x+5)\text{ is also a factor.}\hfill \text{[ICSE 2011]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=x^3+2x^2-kx+10$
$\displaystyle \text{Since }(x-2)\text{ is a factor of }f(x),\text{ by the Factor Theorem, }f(2)=0.$
$\displaystyle 2^3+2(2)^2-k(2)+10=0$
$\displaystyle 8+8-2k+10=0$
$\displaystyle 26-2k=0$
$\displaystyle k=13$
$\displaystyle \text{Now, }f(-5)=(-5)^3+2(-5)^2-13(-5)+10$
$\displaystyle =-125+50+65+10=0$
$\displaystyle \therefore (x+5)\text{ is also a factor of }x^3+2x^2-13x+10.$
$\\$

$\displaystyle \textbf{Question 9: }\text{If }(x-2)\text{ is a factor of }2x^3+ax^2+bx-14\text{ and when the}$
$\displaystyle \text{expression is divided by }(x-3),\text{ it leaves a remainder }52.\text{ Find the value of }a\text{ and }b.\hfill \text{[ICSE 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=2x^3+ax^2+bx-14$
$\displaystyle \text{Since }(x-2)\text{ is a factor of }f(x),\text{ by the Factor Theorem, }f(2)=0.$
$\displaystyle 2(2)^3+a(2)^2+b(2)-14=0$
$\displaystyle 4a+2b=-2$
$\displaystyle 2a+b=-1\qquad\cdots(i)$
$\displaystyle \text{Also, }f(3)=52$
$\displaystyle 2(3)^3+a(3)^2+b(3)-14=52$
$\displaystyle 54+9a+3b-14=52$
$\displaystyle 9a+3b=12$
$\displaystyle 3a+b=4\qquad\cdots(ii)$
$\displaystyle \text{Subtracting }(i)\text{ from }(ii),$
$\displaystyle a=5$
$\displaystyle \text{Substituting }a=5\text{ in }(i),$
$\displaystyle 2(5)+b=-1$
$\displaystyle 10+b=-1$
$\displaystyle b=-11$
$\displaystyle \therefore a=5\text{ and }b=-11.$
$\\$

Question 10: Using remainder theorem, factorize $x^3+10x^2-37x+26$ completely. [ICSE2014]
$\displaystyle \text{Answer:}$
For $x = 1$ ,
Remainder: $= (1)^3+10(1)^2-37(1)+26 = 1+10-37+26=0$
Hence $(x-1)$ is a factor of $x^3+10x^2-37x+26$

$x-1 ) \overline {x^3+10x^2-37x+26} (x^2+11x-26$
$(-) \ \ \underline {x^3-x^2}$
$11x^2-37x+26$
$(-) \ \ \underline{11x^2-11x}$
$-26x+26$
$(-) \ \ \underline{ -26x+26}$
$\times$

$x^3+10x^2-37x+26 = (x-1)(x^2+11x-26)$
$= (x-1)(x^2-2x+13x-26)$
$= (x-1)[x(x-2)+13(x-2)]$
$= (x-1)(x-2)(x+13)$
Hence $x^3+10x^2-37x+26 =(x-1)(x-2)(x+13)$

$\displaystyle \textbf{Question 11: }\text{Find the value of }a,\text{ if the two polynomials }ax^3+3x^2-9\text{ and}$
$\displaystyle 2x^3+4x+a\text{ leave the same remainder when divided by }(x+3).\hfill \text{[ICSE 2015]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }P_1(x)=ax^3+3x^2-9$
$\displaystyle \text{and }P_2(x)=2x^3+4x+a$
$\displaystyle \text{Since both polynomials leave the same remainder when divided by }(x+3),$
$\displaystyle P_1(-3)=P_2(-3)$
$\displaystyle a(-3)^3+3(-3)^2-9=2(-3)^3+4(-3)+a$
$\displaystyle -27a+27-9=-54-12+a$
$\displaystyle -27a+18=-66+a$
$\displaystyle -28a=-84$
$\displaystyle a=3$
$\displaystyle \therefore \text{The value of }a\text{ is }3.$
$\\$

$\displaystyle \textbf{Question 12: }\text{Using Remainder Theorem, find the value of }k,\text{ if on dividing}$
$\displaystyle 2x^3+3x^2-kx+5\text{ by }(x-2),\text{ it leaves a remainder }7.\hfill \text{[ICSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=2x^3+3x^2-kx+5$
$\displaystyle \text{Since }f(x)\text{ is divided by }(x-2)\text{ and leaves remainder }7,$
$\displaystyle f(2)=7$
$\displaystyle 2(2)^3+3(2)^2-k(2)+5=7$
$\displaystyle 16+12-2k+5=7$
$\displaystyle 33-2k=7$
$\displaystyle 2k=26$
$\displaystyle k=13$
$\displaystyle \therefore \text{The value of }k\text{ is }13.$
$\\$

$\displaystyle \textbf{Question 13: }\text{Use Remainder Theorem to factorize the polynomial}$
$\displaystyle 2x^3+3x^2-9x-10.\hfill \text{[ICSE 2018]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=2x^3+3x^2-9x-10$
$\displaystyle f(-1)=2(-1)^3+3(-1)^2-9(-1)-10$
$\displaystyle =-2+3+9-10=0$
$\displaystyle \text{Therefore, by the Factor Theorem, }(x+1)\text{ is a factor of }f(x).$
$\displaystyle \text{Dividing }2x^3+3x^2-9x-10\text{ by }(x+1),\text{ we get }2x^2+x-10.$
$\displaystyle \therefore 2x^3+3x^2-9x-10=(x+1)(2x^2+x-10)$
$\displaystyle =(x+1)(2x^2+5x-4x-10)$
$\displaystyle =(x+1)\{x(2x+5)-2(2x+5)\}$
$\displaystyle =(x+1)(x-2)(2x+5)$
$\displaystyle \therefore \text{The factorized form is }(x+1)(x-2)(2x+5).$
$\\$

$\displaystyle \textbf{Question 14: }\text{Using the Remainder Theorem, find the remainders obtained when}$
$\displaystyle x^3+(kx+8)x+k\text{ is divided by }(x+1)\text{ and }(x-2).$
$\displaystyle \text{Hence find }k,\text{ if the sum of the two remainders is }1.\hfill \text{[ICSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=x^3+(kx+8)x+k$
$\displaystyle \text{The remainder when }f(x)\text{ is divided by }(x+1)\text{ is }f(-1).$
$\displaystyle f(-1)=(-1)^3+\{k(-1)+8\}(-1)+k$
$\displaystyle =-1+(-k+8)(-1)+k$
$\displaystyle =-1+k-8+k=2k-9$
$\displaystyle \text{The remainder when }f(x)\text{ is divided by }(x-2)\text{ is }f(2).$
$\displaystyle f(2)=(2)^3+\{k(2)+8\}(2)+k$
$\displaystyle =8+(2k+8)(2)+k$
$\displaystyle =8+4k+16+k=5k+24$
$\displaystyle \text{According to the given condition,}$
$\displaystyle f(-1)+f(2)=1$
$\displaystyle (2k-9)+(5k+24)=1$
$\displaystyle 7k+15=1$
$\displaystyle 7k=-14$
$\displaystyle k=-2$
$\displaystyle \therefore \text{The required value of }k\text{ is }-2.$
$\\$

$\displaystyle \textbf{Question 15: }\text{What must be added to the polynomial }2x^3-3x^2-8x,$
$\displaystyle \text{so that it leaves a remainder }10\text{ when divided by }(2x+1)?\hfill \text{[ICSE 2020]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=2x^3-3x^2-8x$
$\displaystyle \text{Putting }2x+1=0,\text{ we get }x=-\frac{1}{2}$
$\displaystyle \therefore \text{The remainder is }f\left(-\frac{1}{2}\right)$
$\displaystyle =2\left(-\frac{1}{2}\right)^3-3\left(-\frac{1}{2}\right)^2-8\left(-\frac{1}{2}\right)$
$\displaystyle =2\left(-\frac{1}{8}\right)-3\left(\frac{1}{4}\right)+4$
$\displaystyle =-\frac{1}{4}-\frac{3}{4}+4$
$\displaystyle =-1+4=3$
$\displaystyle \text{But the required remainder is }10.$
$\displaystyle \therefore \text{The required number to be added is }10-3=7.$
$\\$

$\displaystyle \textbf{Question 16: }\text{Find the value of }k,\text{ if }4x^3-2x^2+kx+5\text{ leaves a remainder }$
$\displaystyle -10 \ \text{when divided by }(2x+1).\hfill \text{[ICSE 2020]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }p(x)=4x^3-2x^2+kx+5$
$\displaystyle \text{Putting }2x+1=0,\text{ we get }x=-\frac{1}{2}$
$\displaystyle \text{Since the remainder is }-10,\text{ by the Remainder Theorem,}$
$\displaystyle p\left(-\frac{1}{2}\right)=-10$
$\displaystyle 4\left(-\frac{1}{2}\right)^3-2\left(-\frac{1}{2}\right)^2+k\left(-\frac{1}{2}\right)+5=-10$
$\displaystyle 4\left(-\frac{1}{8}\right)-2\left(\frac{1}{4}\right)-\frac{k}{2}+5=-10$
$\displaystyle -\frac{1}{2}-\frac{1}{2}-\frac{k}{2}+5=-10$
$\displaystyle 4-\frac{k}{2}=-10$
$\displaystyle -\frac{k}{2}=-14$
$\displaystyle k=28$
$\displaystyle \therefore \text{The value of }k\text{ is }28.$
$\\$

$\displaystyle \textbf{Question 17: }\text{The polynomial }x^3-2x^2+ax+12\text{ when divided by }(x+1)\text{ leaves remainder }20.$
$\displaystyle \text{The value of }a\text{ is equal to}\hfill \text{[ICSE Semester I 2022]}$
$\displaystyle \text{(a) }-31\qquad\text{(b) }9\qquad\text{(c) }11\qquad\text{(d) }-11$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(d) Let }f(x)=x^3-2x^2+ax+12$
$\displaystyle \text{Since }f(x)\text{ is divided by }(x+1)\text{ and leaves remainder }20,$
$\displaystyle f(-1)=20$
$\displaystyle (-1)^3-2(-1)^2+a(-1)+12=20$
$\displaystyle -1-2-a+12=20$
$\displaystyle -a+9=20$
$\displaystyle -a=11$
$\displaystyle a=-11$
$\displaystyle \therefore \text{Option (d) is correct.}$
$\\$

$\displaystyle \textbf{Question 18: }\text{If the polynomial }2x^2-7x-1\text{ is divided by }(x+3),\text{ then the remainder is}$
$\displaystyle \hfill \text{[ICSE Semester I 2022]}$
$\displaystyle \text{(a) }-4\qquad\text{(b) }38\qquad\text{(c) }-3\qquad\text{(d) }2$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(b) Let }f(x)=2x^2-7x-1$
$\displaystyle \text{Since }f(x)\text{ is divided by }(x+3),\text{ by the Remainder Theorem, the remainder is }f(-3).$
$\displaystyle f(-3)=2(-3)^2-7(-3)-1$
$\displaystyle =18+21-1=38$
$\displaystyle \therefore \text{Option (b) is correct.}$
$\\$

$\displaystyle \textbf{Question 19: }\text{Factorise the given polynomial completely, using Remainder Theorem}$
$\displaystyle 6x^3+25x^2+31x+10.\hfill \text{[ICSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=6x^3+25x^2+31x+10$
$\displaystyle f(-2)=6(-2)^3+25(-2)^2+31(-2)+10$
$\displaystyle =-48+100-62+10=0$
$\displaystyle \text{Therefore, by the Factor Theorem, }(x+2)\text{ is a factor of }f(x).$
$\displaystyle \text{Dividing }6x^3+25x^2+31x+10\text{ by }(x+2),\text{ we get }6x^2+13x+5.$
$\displaystyle \therefore 6x^3+25x^2+31x+10=(x+2)(6x^2+13x+5)$
$\displaystyle =(x+2)(6x^2+10x+3x+5)$
$\displaystyle =(x+2)\{2x(3x+5)+1(3x+5)\}$
$\displaystyle =(x+2)(2x+1)(3x+5)$
$\displaystyle \therefore \text{The factorized form is }(x+2)(2x+1)(3x+5).$
$\\$