Class 10 – Factorization – ICSE Board Problems (1) – ICSE / ISC / CBSE Mathematics Portal for K12 Students

$\displaystyle \textbf{Question 1: }\text{Find the remainder, when }2x^3 - 3x^2 + 7x - 8\text{ is divided by }(x-1). \hspace{3.0cm} \text{[ICSE 2000]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=2x^3-3x^2+7x-8$
$\displaystyle \text{Since, }f(x)\text{ is divided by }x-1,\text{ therefore we have to find the value of }f(1).$
$\displaystyle f(1)=2(1)^3-3(1)^2+7(1)-8$
$\displaystyle =2-3+7-8=9-11=-2$
$\displaystyle \text{Hence, the required remainder is }-2.$

$\displaystyle \textbf{Question 2: }\text{A function }f\text{ is defined by }f(x) = 144 - 16x^2,\text{ calculate }f(2).$
$\displaystyle \text{Also, find the value of }x,\text{ when }f(x) = 0. \hspace{3.0cm} \text{[ICSE 2001]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }f(x)=144-16x^2$
$\displaystyle \text{Put }x=2,\text{ we get}$
$\displaystyle f(2)=144-16(2)^2$
$\displaystyle =144-16\times4$
$\displaystyle =144-64=80$
$\displaystyle \text{Also, given }f(x)=0$
$\displaystyle 144-16x^2=0$
$\displaystyle 16x^2=144$
$\displaystyle x^2=\frac{144}{16}$
$\displaystyle x^2=9$
$\displaystyle x=\pm3$

Question 3: Find the value of $a$ , if $(x-a)$ is a factor of $x^3-ax^2+x+2$ . [ICSE2003]
$\displaystyle \text{Answer:}$
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 4: If $x^3+ax^2+bx+6$ has $(x-2)$ as a factor and leaves a remainder of $3$ when divided by $(x-3)$ , find the value of $a \ and \ b$ . [ICSE2005]
$\displaystyle \text{Answer:}$
When $x=2$ , Remainder $= 0$
$\Rightarrow (2)^3+a(2)^2+b(2)+6 = 0$
$\Rightarrow 4a+2b = - 14$ … … … … … i)
When $x = 3$ , Remainder $= 3$
$\Rightarrow (3)^3+a(3)^2+b(3)+6 = 3$
$\Rightarrow 9a+3b=-30$ … … … … … ii)
Solving i) and ii) $a = -3 \ and \ b = -1$

Question 5: When divided by $(x-3)$ the polynomials $x^3-px^2+x+6$ and $2x^3-x^2-(p+3)x-6$ leave the same remainder. Find the value of $p$ . [ICSE2010]
$\displaystyle \text{Answer:}$
When $x=3$
$\text{Remainder}_1 = (3)^3-p(3)^2+(3)+6$
$= 27-9p+9$
$= 36-9p$
$\text{Remainder}_2 = 2(3)^3-(3)^2-(p+3)(3)-6$
$=54-9-3p-9-6$
$=30-39$
$\text{Given Remainder}_1 = \text{Remainder}_1$
$36-9p=30-3p$
$6=6p \Rightarrow p =1$

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

$\displaystyle \textbf{Question 7: }\text{When divided by }x - 3,\text{ the polynomials}$
$\displaystyle x^3 - px^2 + x + 6\text{ and }2x^3 - x^2 - (p + 3)x - 6\text{ leave the same remainder.}$
$\displaystyle \text{Find the value of }p. \hspace{3.0cm} \text{[ICSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=x^3-px^2+x+6$
$\displaystyle \text{and }g(x)=2x^3-x^2-(p+3)x-6$
$\displaystyle \text{Since, the given polynomials leave the same remainder, when divided by }x-3.$
$\displaystyle \therefore f(3)=g(3)$
$\displaystyle 3^3-p(3)^2+3+6=2(3)^3-(3)^2-(p+3)3-6$
$\displaystyle 27-9p+3+6=54-9-3(p+3)-6$
$\displaystyle 36-9p=39-3(p+3)$
$\displaystyle 36-9p=39-3p-9$
$\displaystyle 36-9p=30-3p$
$\displaystyle 36-30=-3p+9p$
$\displaystyle 6p=6$
$\displaystyle p=1$
$\displaystyle \text{Hence, the value of }p\text{ is }1.$

Question 8: Find the value of $k$ if $(x-2)$ is a factor of $x^3+2x^2-kx+10$ . Hence determine whether $(x+5)$ is also a factor. [ICSE2011]
$\displaystyle \text{Answer:}$
Let $f(x) = x^3+2x^2-kx+10$ .
Since given that $(x-2)$ is a factor $f(2) = 0$
Substituting the value of $x =2$ in the above function we get:
$f(2) = 0$
$f(2) = 8+8-2k+10=0$
$\Rightarrow k=13$
For $(x + 5)$ to be a factor $f(-5) = 0$
Substituting the value of $x =-5$ in the above function we get:
$f(-5) = (-5)^3+2(-5)^2-k(-5)+10 = -125+50+65+10=0$
Hence $(x+5$ ) is a factor of $x^3+2x^2-kx+10$

Question 9: If $(x-2)$ is a factor of the expression $2x^3+ax^2+bx-14$ and when the expression is divided by $(x-3)$ , it leaves a remainder $52$ . Find the value of $a \ and \ b$ . [ICSE2013]
$\displaystyle \text{Answer:}$
When $x=2$ , Remainder $= 0$
$\Rightarrow 2(2)^3+a(2)^2+b(2)-14=0$
$\Rightarrow 4a+2b=02$
$\Rightarrow 2a+b=-1$ … … … … … i)
When $x = 3$ , Remainder $= 52$
$\Rightarrow 2(3)^3+a(3)^2+b(3)-14=52$
$\Rightarrow 9a+3b=12$
$\Rightarrow 3a+b=4$ … … … … … ii)
Solving i) and ii), we get $a = 5 \ and \ b = -11$

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 $x^3+10x^2-37x+26 =(x-1)(x-2)(x+13)$

$\displaystyle \textbf{Question 11: }\text{Find the value of }a,\text{ if the two polynomials} ax^3 + 3x^2 - 9\text{ and } \\ 2x^3 + 4x + a\text{ leave the same remainder when divided by }x + 3. \hspace{3.0cm} \text{[ICSE 2015]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let given two polynomials are}$
$\displaystyle P_1(x)=ax^3+3x^2-9$
$\displaystyle \text{and }P_2(x)=2x^3+4x+a$
$\displaystyle \text{If both are divided by }x+3\text{ and having same remainder, then }P_1(-3)=P_2(-3)$
$\displaystyle a(-3)^3+3(-3)^2-9$
$\displaystyle =2(-3)^3+4(-3)+a$
$\displaystyle -27a+27-9=-54-12+a$
$\displaystyle -28a=-66-18$
$\displaystyle -28a=-84$
$\displaystyle a=\frac{-84}{-28}$
$\displaystyle a=3$
$\displaystyle \text{Hence, the value of }a\text{ is }3.$

$\displaystyle \textbf{Question 12: }\text{Using remainder theorem, find the value of }k,\text{ if on dividing}$
$\displaystyle 2x^3 + 3x^2 - kx + 5\text{ by }x - 2,\text{ leaves a remainder }7. \hspace{3.0cm} \text{[ICSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=2x^3+3x^2-kx+5$
$\displaystyle \text{Since, }f(x)\text{ is divided by }(x-2)\text{ leaves remainder }7.$
$\displaystyle \therefore f(2)=7$
$\displaystyle 2(2)^3+3(2)^2-k\times2+5=7$
$\displaystyle 16+12-2k+5=7$
$\displaystyle 33-2k=7$
$\displaystyle 2k=33-7$
$\displaystyle 2k=26$
$\displaystyle k=13$

$\displaystyle \textbf{Question 13: }\text{Use remainder theorem to factorize the polynomial}$
$\displaystyle 2x^3 + 3x^2 - 9x - 10. \hspace{3.0cm} \text{[ICSE 2018]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=2x^3+3x^2-9x-10$
$\displaystyle \text{For }x=-1,$
$\displaystyle f(x)=f(-1)$
$\displaystyle =2(-1)^3+3(-1)^2-9(-1)-10$
$\displaystyle =-2+3+9-10$
$\displaystyle =12-12=0$
$\displaystyle \text{Hence, }(x+1)\text{ is a factor of }f(x).$
$\displaystyle \therefore 2x^3+3x^2-9x-10$
$\displaystyle =(x+1)(2x^2+x-10)$
$\displaystyle =(x+1)(2x^2+5x-4x-10)$
$\displaystyle =(x+1)\{x(2x+5)-2(2x+5)\}$
$\displaystyle =(x+1)(x-2)(2x+5)$

$\displaystyle \textbf{Question 14: }\text{Using the remainder theorem find the remainders obtained when}$
$\displaystyle x^3 + (kx + 8)x + k\text{ is divided by }x + 1\text{ and }x - 2.$
$\displaystyle \text{Hence find }k,\text{ if the sum of the two remainders is }1. \hspace{3.0cm} \text{[ICSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=x^3+(kx+8)x+k$
$\displaystyle \text{Then, the remainder, when }f(x)\text{ is divided by }(x+1)\text{ and }(x-2)\text{ respectively is given by}$
$\displaystyle f(-1)=(-1)^3+[k(-1)+8](-1)+k$
$\displaystyle =-1+(-k+8)(-1)+k$
$\displaystyle =-1+k-8+k=2k-9$
$\displaystyle \text{and }f(2)=(2)^3+(k\times2+8)2+k$
$\displaystyle =8+(4k+16)+k$
$\displaystyle =5k+24$
$\displaystyle \text{According to the given condition,}$
$\displaystyle f(-1)+f(2)=1$
$\displaystyle 2k-9+5k+24=1$
$\displaystyle 7k+15=1$
$\displaystyle 7k=-14$
$\displaystyle k=-2$

$\displaystyle \textbf{Question 15: }\text{What must be added to the polynomial }2x^3 - 3x^2 - 8x,$
$\displaystyle \text{so that it leaves a remainder }10\text{ when divided by }2x + 1? \hspace{3.0cm} \text{[ICSE 2020]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=2x^3-3x^2-8x$
$\displaystyle \text{Put }2x+1=0\Rightarrow x=-\frac{1}{2}$
$\displaystyle \therefore \text{Remainder}=f\left(-\frac{1}{2}\right)$
$\displaystyle =2\left(-\frac{1}{2}\right)^3-3\left(-\frac{1}{2}\right)^2-8\left(-\frac{1}{2}\right)$
$\displaystyle =2\left(-\frac{1}{8}\right)-3\left(\frac{1}{4}\right)+4$
$\displaystyle =-\frac{1}{4}-\frac{3}{4}+4$
$\displaystyle =-1+4=3$
$\displaystyle \text{But remainder }=10\text{ (given)}$
$\displaystyle \text{Hence, the required number added is }10-3=7$

$\displaystyle \textbf{Question 16: }\text{Find the value of }k,\text{ if }4x^3 - 2x^2 + kx + 5\text{ leaves remainder }-10 \text{when divided by }2x + 1. \hspace{3.0cm} \text{[ICSE 2020]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }p(x)=4x^3-2x^2+kx+5\text{ and }g(x)=2x+1.$
$\displaystyle \text{Put }g(x)=0\Rightarrow 2x+1=0$
$\displaystyle \Rightarrow x=-\frac{1}{2}$
$\displaystyle \text{According to the question, when }p(x)\text{ is divided by }g(x),\text{ then remainder is }-10.$
$\displaystyle \therefore p\left(-\frac{1}{2}\right)=-10$
$\displaystyle 4\left(-\frac{1}{2}\right)^3-2\left(-\frac{1}{2}\right)^2+k\left(-\frac{1}{2}\right)+5=-10$
$\displaystyle =4\left(-\frac{1}{8}\right)-2\left(\frac{1}{4}\right)-\frac{k}{2}+5=-10$
$\displaystyle =-\frac{4}{8}-\frac{2}{4}-\frac{k}{2}+5=-10$
$\displaystyle =-1-\frac{k}{2}+5=-10$
$\displaystyle \Rightarrow -\frac{k}{2}=14$
$\displaystyle \Rightarrow k=28$

$\displaystyle \textbf{Question 17: }\text{The polynomial }x^3 - 2x^2 + ax + 12\text{ when divided by }(x+1)\text{ leaves remainder }20.$
$\displaystyle \text{The value of }a\text{ is equal to} \hspace{3.0cm} \text{[ICSE Semester I 2022]}$
$\displaystyle (a)\ -31 \qquad (b)\ 9 \qquad (c)\ 11 \qquad (d)\ -11$
$\displaystyle \text{Answer:}$
$\displaystyle (d)\ \text{Let }f(x)=x^3-2x^2+ax+12$
$\displaystyle \text{Since, }f(x)\text{ is divided by }(x+1)\text{ leaves remainder }20.$
$\displaystyle \therefore f(-1)=20$
$\displaystyle (-1)^3-2(-1)^2+a(-1)+12=20$
$\displaystyle -1-2-a+12=20$
$\displaystyle -3-a+12=20$
$\displaystyle -a+9=20\Rightarrow -a=11$
$\displaystyle a=-11$

$\displaystyle \textbf{Question 18: }\text{If a polynomial }2x^2 - 7x - 1\text{ is divided by }(x+3),\text{ then the remainder is}$
$\displaystyle \text{[ICSE Semester I 2022]}$
$\displaystyle (a)\ -4 \qquad (b)\ 38 \qquad (c)\ -3 \qquad (d)\ 2$
$\displaystyle \text{Answer:}$
$\displaystyle (b)\ \text{Given, polynomial is }2x^2-7x-1.$
$\displaystyle \text{To divide }2x^2-7x-1\text{ by }(x+3).$
$\displaystyle \text{Let }f(x)=2x^2-7x-1\text{ and put }x=-3$
$\displaystyle f(-3)=2(-3)^2-7(-3)-1$
$\displaystyle =18+21-1=38$
$\displaystyle \text{Hence, remainder is }38.$

$\displaystyle \textbf{Question 19: }\text{Factorise the given polynomial completely, using remainder theorem}$
$\displaystyle 6x^3 + 25x^2 + 31x + 10. \hspace{3.0cm}\text{[ICSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, polynomial is}$
$\displaystyle 6x^3+25x^2+31x+10$
$\displaystyle \text{Let }f(x)=6x^3+25x^2+31x+10$
$\displaystyle \text{For }x=-2$
$\displaystyle f(-2)=6(-2)^3+25(-2)^2+31(-2)+10$
$\displaystyle =-48+100-62+10$
$\displaystyle =110-110=0$
$\displaystyle \text{Hence, }(x+2)\text{ is a factor of }6x^3+25x^2+31x+10.$
$\displaystyle \therefore 6x^3+25x^2+31x+10$
$\displaystyle =(x+2)(6x^2+13x+5)$
$\displaystyle \text{Now, let }g(x)=6x^2+13x+5$
$\displaystyle \text{On putting }x=-\frac{1}{2},\text{ we get}$
$\displaystyle g\left(-\frac{1}{2}\right)=6\left(-\frac{1}{2}\right)^2+13\left(-\frac{1}{2}\right)+5$
$\displaystyle =6\left(\frac{1}{4}\right)-\frac{13}{2}+5$
$\displaystyle =\frac{6-26+20}{4}=0$
$\displaystyle \text{Hence, }(2x+1)\text{ is a factor of }6x^2+13x+5.$
$\displaystyle \therefore g(x)=(2x+1)(3x+5)$
$\displaystyle \text{Hence, }6x^3+25x^2+31x+10$
$\displaystyle =(x+2)(2x+1)(3x+5)$