Class 10 – Factorization – ICSE Board Problems (2) – ICSE / ISC / CBSE Mathematics Portal for K12 Students

$\displaystyle \textbf{Question 1: }\text{Find the value of the constants }a\text{ and }b,\text{ if }(x - 2)\text{ and }(x + 3)\text{ are both} \\ \text{factors of the expression }x^3 + ax^2 + bx - 12.\ \text{[ICSE 2001]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }p(x)=x^3+ax^2+bx-12\ \ \text{(i)}$
$\displaystyle \text{Since, }(x-2)\text{ and }(x+3)\text{ are the factors of }p(x).$
$\displaystyle p(2)=0$
$\displaystyle (2)^3+a(2)^2+b(2)-12=0$
$\displaystyle 8+4a+2b-12=0$
$\displaystyle 4a+2b-4=0$
$\displaystyle 2a+b-2=0\ \ \text{(ii)}$
$\displaystyle \text{and }p(-3)=0$
$\displaystyle (-3)^3+a(-3)^2+b(-3)-12=0$
$\displaystyle -27+9a-3b-12=0$
$\displaystyle 9a-3b-39=0$
$\displaystyle 3a-b-13=0\ \ \text{(iii)}$
$\displaystyle \text{On adding Eqs.\ (ii) and (iii), we get}$
$\displaystyle (2a+b-2)+(3a-b-13)=0$
$\displaystyle 5a-15=0$
$\displaystyle 5a=15$
$\displaystyle a=3$
$\displaystyle \text{On putting }a=3\text{ in Eq.\ (ii), we get}$
$\displaystyle 2(3)+b-2=0$
$\displaystyle 6+b-2=0$
$\displaystyle b+4=0$
$\displaystyle b=-4$
$\displaystyle \text{Hence, the values of }a\text{ and }b\text{ are }3\text{ and }-4.$

$\displaystyle \textbf{Question 2: }\text{If }(x - 2)\text{ is a factor of }2x^3 - x^2 - px - 2.$
$\displaystyle \text{(i) Find the value of }p.$
$\displaystyle \text{(ii) With the value of }p,\text{ factorise the above expression completely.}\ \text{[ICSE 2008]}$
$\displaystyle \text{Answer:}$
$\displaystyle (i)\ \text{Let }f(x)=2x^3-x^2-px-2\ \ \text{(i)}$
$\displaystyle \text{Since, }(x-2)\text{ is a factor of }f(x).$
$\displaystyle f(2)=0$
$\displaystyle 2(2)^3-(2)^2-p(2)-2=0$
$\displaystyle 16-4-2p-2=0$
$\displaystyle 10-2p=0$
$\displaystyle -2p=-10$
$\displaystyle 2p=10$
$\displaystyle p=5$
$\displaystyle (ii)\ \text{On putting }p=5\text{ in Eq.\ (i), we get}$
$\displaystyle f(x)=2x^3-x^2-5x-2$
$\displaystyle \text{Since, }(x-2)\text{ is a factor of }f(x),\text{ so we divide }f(x)\text{ by }(x-2)\text{ using long division method.}$
$\displaystyle f(x)=(x-2)(2x^2+3x+1)$

$\displaystyle \textbf{Question 3: }\text{ }(x - 2)\text{ is a factor of the expression }x^3 + ax^2 + bx + 6.\text{ When this expression} \\ \text{is divided by }(x - 3),\text{ it leaves the remainder }3.\text{ Find the values of }a\text{ and }b.\ \text{[ICSE 2005]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }p(x)=x^3+ax^2+bx+6\ \ \text{(i)}$
$\displaystyle \text{Since, }(x-2)\text{ is a factor of }p(x).$
$\displaystyle p(2)=0$
$\displaystyle (2)^3+a(2)^2+b(2)+6=0$
$\displaystyle 8+4a+2b+6=0$
$\displaystyle 4a+2b+14=0$
$\displaystyle 2a+b+7=0\ \ \text{(ii)}$
$\displaystyle \text{Also, when }p(x)\text{ is divided by }(x-3),\text{ it leaves the remainder }3.$
$\displaystyle p(3)=3$
$\displaystyle (3)^3+a(3)^2+b(3)+6=3$
$\displaystyle 27+9a+3b+6=3$
$\displaystyle 9a+3b+30=0$
$\displaystyle 3a+b+10=0\ \ \text{(iii)}$
$\displaystyle \text{On subtracting Eq.\ (ii) from Eq.\ (iii), we get}$
$\displaystyle (3a+b+10)-(2a+b+7)=0$
$\displaystyle a+3=0$
$\displaystyle a=-3$
$\displaystyle \text{On putting }a=-3\text{ in Eq.\ (iii), we get}$
$\displaystyle 3(-3)+b+10=0$
$\displaystyle -9+b+10=0$
$\displaystyle b=-1$
$\displaystyle \text{Hence, }a=-3\text{ and }b=-1.$

$\displaystyle \textbf{Question 4: }\text{Show that }2x + 7\text{ is a factor of }2x^3 + 5x^2 - 11x - 14.\text{ Hence, factorise the given expression} \\ \text{completely, using factor theorem.}\ \text{[ICSE 2006]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }p(x)=2x^3+5x^2-11x-14$
$\displaystyle \text{On comparing }2x+7\text{ with }(x-a),\text{ we get }a=-\frac{7}{2}$
$\displaystyle \text{Then, }p\left(\frac{7}{2}\right)=2\left(-\frac{7}{2}\right)^3+5\left(-\frac{7}{2}\right)^2-11\left(-\frac{7}{2}\right)-14$
$\displaystyle =-\frac{343}{8}+5\times\frac{49}{4}+\frac{77}{2}-14$
$\displaystyle =-\frac{686}{8}+\frac{245}{4}+\frac{77}{2}-14$
$\displaystyle =-\frac{686+490+308-112}{8}$
$\displaystyle =-\frac{798+798}{8}=0$
$\displaystyle \text{Hence, }(2x+7)\text{ is one of the factors of }p(x)=2x^3+5x^2-11x-14\ \text{(by factor theorem)}$
$\displaystyle \text{Now, we divide }p(x)\text{ by }2x+7,\text{ using long division method.}$
$\displaystyle p(x)=(2x+7)(x^2-x-2)$
$\displaystyle =(2x+7)[x^2-2x+x-2]$
$\displaystyle =(2x+7)[x(x-2)+1(x-2)]$
$\displaystyle =(2x+7)(x-2)(x+1)$

$\displaystyle \textbf{Question 5: }\text{Show that }(x - 1)\text{ is a factor of }x^3 - 7x^2 + 14x - 8.\text{ Hence, completely factorise} \\ \text{the above expression.}\ \text{[ICSE 2007]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=x^3-7x^2+14x-8\ \ \text{(i)}$
$\displaystyle \text{Consider, }x-1=0\Rightarrow x=1$
$\displaystyle \text{Put }x=1\text{ in Eq.\ (i), we get}$
$\displaystyle f(1)=(1)^3-7(1)^2+14\times1-8$
$\displaystyle =1-7+14-8=15-15=0$
$\displaystyle \text{Hence, }(x-1)\text{ is a factor of }f(x).$
$\displaystyle \text{Now, we divide }f(x)\text{ by }(x-1)\text{ using long division method.}$
$\displaystyle f(x)=(x-1)(x^2-6x+8)$
$\displaystyle =(x-1)(x^2-4x-2x+8)$
$\displaystyle =(x-1)[x(x-4)-2(x-4)]$
$\displaystyle =(x-1)(x-2)(x-4)$
$\displaystyle 18.\ \text{Let }p(x)=2x^3+5x^2-11x-14$
$\displaystyle \text{On comparing }2x+7\text{ with }(x-a),\text{ we get }a=-\frac{7}{2}$
$\displaystyle \text{Then, }p\left(\frac{7}{2}\right)=2\left(-\frac{7}{2}\right)^3+5\left(-\frac{7}{2}\right)^2-11\left(-\frac{7}{2}\right)-14$
$\displaystyle =-\frac{343}{8}+5\times\frac{49}{4}+\frac{77}{2}-14$
$\displaystyle =-\frac{686}{8}+\frac{245}{4}+\frac{77}{2}-14$
$\displaystyle =-\frac{686+490+308-112}{8}$
$\displaystyle =-\frac{798+798}{8}=0$
$\displaystyle \text{Hence, }(2x+7)\text{ is one of the factors of }p(x)=2x^3+5x^2-11x-14\ \text{(by factor theorem)}$
$\displaystyle \text{Now, we divide }p(x)\text{ by }2x+7,\text{ using long division method.}$
$\displaystyle p(x)=(2x+7)(x^2-x-2)$

$\displaystyle \textbf{Question 6: }\text{Given that }x + 2\text{ and }x + 3\text{ are factors of }2x^3 + ax^2 + 7x - b.\text{ Determine the values of }a\text{ and }b.\ \text{[ICSE 2009]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=2x^3+ax^2+7x-b$
$\displaystyle \text{Since, }(x+2)\text{ is a factor of }f(x),\text{ therefore }f(-2)=0$
$\displaystyle 2(-2)^3+a(-2)^2+7(-2)-b=0$
$\displaystyle -16+4a-14-b=0$
$\displaystyle 4a-b-30=0\ \ \text{(i)}$
$\displaystyle \text{Also, }(x+3)\text{ is a factor of }f(x),\text{ therefore }f(-3)=0$
$\displaystyle 2(-3)^3+a(-3)^2+7(-3)-b=0$
$\displaystyle -54+9a-21-b=0$
$\displaystyle 9a-b-75=0\ \ \text{(ii)}$
$\displaystyle \text{Subtract Eq.\ (i) from Eq.\ (ii), we get}$
$\displaystyle (9a-b-75)-(4a-b-30)=0$
$\displaystyle 5a-45=0$
$\displaystyle 5a=45$
$\displaystyle a=9$
$\displaystyle \text{Put the value of }a\text{ in Eq.\ (i), we get}$
$\displaystyle 4(9)-b-30=0$
$\displaystyle 36-b-30=0$
$\displaystyle 6-b=0$
$\displaystyle b=6$
$\displaystyle \text{Hence, }a=9\text{ and }b=6.$

$\displaystyle \textbf{Question 7: }\text{Use the factor theorem to factorise the following expression }2x^3 + x^2 - 13x + 6.\ \text{[ICSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=2x^3+x^2-13x+6$
$\displaystyle \text{Using hit and trial method,}$
$\displaystyle \text{Put }x=1,\text{ we get}$
$\displaystyle f(1)=2(1)^3+(1)^2-13(1)+6$
$\displaystyle =2+1-13+6=-4\neq 0$
$\displaystyle \text{Put }x=-1,\text{ we get}$
$\displaystyle f(-1)=2(-1)^3+(-1)^2-13(-1)+6$
$\displaystyle =-2+1+13+6=18\neq 0$
$\displaystyle \text{Put }x=2,\text{ we get}$
$\displaystyle f(2)=2(2)^3+(2)^2-13(2)+6$
$\displaystyle =2\times8+4-26+6=16-16=0$
$\displaystyle \text{So, }(x-2)\text{ is one of the factors of }f(x).$
$\displaystyle \text{Now, we divide }f(x)\text{ by }x-2,\text{ using long division method.}$
$\displaystyle f(x)=(x-2)(2x^2+5x-3)\ \ \text{(i)}$
$\displaystyle \text{Now, to find the factors of }2x^2+5x-3.$
$\displaystyle 2x^2+5x-3=2x^2+6x-x-3$
$\displaystyle =2x(x+3)-1(x+3)$
$\displaystyle =(2x-1)(x+3)$
$\displaystyle \therefore \text{Eq.\ (i) can be rewritten as }f(x)=(x-2)(2x-1)(x+3)$

$\displaystyle \textbf{Question 8: }\text{Find the value of }k,\text{ if }(x - 2)\text{ is a factor of }x^3 + 2x^2 - kx + 10.\text{ Hence, determine whether }(x + 5)\text{ is also a factor.}\ \text{[ICSE 2011]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Firstly, consider }f(x)=x^3+2x^2-kx+10$
$\displaystyle \text{Since, }(x-2)\text{ is a factor of }f(x).$
$\displaystyle f(2)=0$
$\displaystyle (2)^3+2(2)^2-k(2)+10=0$
$\displaystyle 8+8-2k+10=0$
$\displaystyle 2k=26$
$\displaystyle k=13$
$\displaystyle \text{On putting }k=13,\text{ we get}$
$\displaystyle f(x)=x^3+2x^2-13x+10$
$\displaystyle \text{Now, to check whether }(x+5)\text{ is a factor of }f(x)\text{ or not.}$
$\displaystyle f(-5)=(-5)^3+2(-5)^2-13(-5)+10$
$\displaystyle =-125+50+65+10=0$
$\displaystyle \therefore (x+5)\text{ is also a factor of }f(x).$

$\displaystyle \textbf{Question 9: }\text{Using the factor theorem, factorise completely the following polynomial } \\ 3x^3 + 2x^2 - 19x + 6.\ \text{[ICSE 2012]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=3x^3+2x^2-19x+6$
$\displaystyle \text{Using hit and trial method,}$
$\displaystyle \text{Put }x=1,\text{ we get}$
$\displaystyle f(1)=3(1)^3+2(1)^2-19(1)+6$
$\displaystyle =3+2-19+6=-8\neq 0$
$\displaystyle \text{Put }x=-1,\text{ we get}$
$\displaystyle f(-1)=3(-1)^3+2(-1)^2-19(-1)+6$
$\displaystyle =-3+2+19+6=24\neq 0$
$\displaystyle \text{Put }x=2,\text{ we get}$
$\displaystyle f(2)=3(2)^3+2(2)^2-19(2)+6$
$\displaystyle =24+8-38+6=38-38=0$
$\displaystyle \text{So, }(x-2)\text{ is a factor of }f(x).$
$\displaystyle \text{Now, we divide }f(x)\text{ by }(x-2)\text{ using long division method.}$
$\displaystyle f(x)=(x-2)(3x^2+8x-3)\ \ \text{(i)}$
$\displaystyle \text{Now, factorise }3x^2+8x-3$
$\displaystyle =3x^2+9x-x-3$
$\displaystyle =3x(x+3)-1(x+3)$
$\displaystyle =(3x-1)(x+3)$
$\displaystyle \therefore f(x)=(x-2)(3x-1)(x+3).$
$\displaystyle =(3x-1)(x+3)$
$\displaystyle \text{From Eq.\ (i), we get}$
$\displaystyle f(x)=(x-2)(3x-1)(x+3)$

$\displaystyle \textbf{Question 10: }\text{If }(x - 2)\text{ is a factor of the expression }2x^3 + ax^2 + bx - 14\text{ and when the expression} \\ \text{is divided by }(x - 3),\text{ it leaves a remainder }52.\text{ Find the values of }a\text{ and }b.\ \text{[ICSE 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=2x^3+ax^2+bx-14\ \ \text{(i)}$
$\displaystyle \text{Since, }(x-2)\text{ is a factor of }f(x).$
$\displaystyle f(2)=0$
$\displaystyle 2(2)^3+a(2)^2+b(2)-14=0$
$\displaystyle 16+4a+2b-14=0$
$\displaystyle 4a+2b+2=0$
$\displaystyle 4a+2b=-2$
$\displaystyle 2a+b=-1\ \ \text{(ii)}$
$\displaystyle \text{Also, given that when }f(x)\text{ is divided by }(x-3),\text{ the remainder is }52.$
$\displaystyle f(3)=52$
$\displaystyle 2(3)^3+a(3)^2+b(3)-14=52$
$\displaystyle 54+9a+3b-14=52$
$\displaystyle 9a+3b=12$
$\displaystyle 3a+b=4\ \ \text{(iii)}$
$\displaystyle \text{On subtracting Eq.\ (iii) from Eq.\ (ii), we get}$
$\displaystyle (2a+b)-(3a+b)=-1-4$
$\displaystyle -a=-5$
$\displaystyle a=5$
$\displaystyle \text{On putting }a=5\text{ in Eq.\ (ii), we get}$
$\displaystyle 2(5)+b=-1$
$\displaystyle 10+b=-1$
$\displaystyle b=-11$
$\displaystyle \text{Hence, }a=5\text{ and }b=-11.$

$\displaystyle \textbf{Question 11: }\text{Using the remainder and factor theorem, factorise the following polynomial } \\ x^3 + 10x^2 - 37x + 26.\ \text{[ICSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=x^3+10x^2-37x+26$
$\displaystyle \text{Using hit and trial method, put }x=1,\text{ we get}$
$\displaystyle f(1)=1^3+10(1)^2-37(1)+26$
$\displaystyle =1+10-37+26=0$
$\displaystyle \therefore (x-1)\text{ is a factor of }f(x).$
$\displaystyle \text{Now, we divide }f(x)\text{ by }(x-1)\text{ using long division method.}$
$\displaystyle \therefore f(x)=(x-1)(x^2+11x-26)\ \ \text{(ii)}$
$\displaystyle \text{Now, factorise }x^2+11x-26$
$\displaystyle =x^2+13x-2x-26$
$\displaystyle =x(x+13)-2(x+13)$
$\displaystyle =(x-2)(x+13)$
$\displaystyle \therefore f(x)=(x-1)(x-2)(x+13)$

$\displaystyle \textbf{Question 12: }\text{Find the value of }a\text{ and }b,\text{ if }x - 1\text{ and }x - 2\text{ are factors of } \\ x^3 - ax + b.\ \text{[ICSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=x^3-ax+b$
$\displaystyle \therefore (x-1)\text{ is a factor of }f(x).$
$\displaystyle f(1)=0\Rightarrow (1)^3-a(1)+b=0$
$\displaystyle 1-a+b=0\Rightarrow a-b=1\ \ \text{(i)}$
$\displaystyle \text{and }(x-2)\text{ is a factor of }f(x).$
$\displaystyle f(2)=0$
$\displaystyle (2)^3-a(2)+b=0$
$\displaystyle 8-2a+b=0$
$\displaystyle 2a-b=8\ \ \text{(ii)}$
$\displaystyle \text{Subtracting Eq.\ (ii) by Eq.\ (i), we get}$
$\displaystyle (2a-b)-(a-b)=8-1$
$\displaystyle a=7$
$\displaystyle \text{Put the value of }a\text{ in Eq.\ (i), we get}$
$\displaystyle 7-b=1$
$\displaystyle b=6$
$\displaystyle \text{Hence, the value of }a=7\text{ and }b=6.$

$\displaystyle \textbf{Question 13:} \text{What must be subtracted from }16x^3 - 8x^2 + 4x + 7\text{ so that the resulting} \\ \text{expression has }2x + 1\text{ as a factor?}\ \text{[ICSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=16x^3-8x^2+4x+7\text{ and }g(x)=2x+1$
$\displaystyle \text{By remainder theorem, when }f(x)\text{ is divided by }g(x)=2x+1,\text{ then remainder is equal to }f\left(-\frac{1}{2}\right)$
$\displaystyle f\left(-\frac{1}{2}\right)=16\left(-\frac{1}{2}\right)^3-8\left(-\frac{1}{2}\right)^2+4\left(-\frac{1}{2}\right)+7$
$\displaystyle =16\left(-\frac{1}{8}\right)-8\left(\frac{1}{4}\right)-2+7$
$\displaystyle =-2-2-2+7=1$
$\displaystyle \therefore 1\text{ must be subtracted from }16x^3-8x^2+4x+7\text{ for getting }(2x+1)\text{ as a factor.}$

$\displaystyle \textbf{Question 14: }\text{If }(x + 2)\text{ and }(x + 3)\text{ are factors of }x^3 + ax + b,\text{ find the values of }a\text{ and }b.\ \text{[ICSE 2018]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=x^3+ax+b$
$\displaystyle \text{Since, }(x+2)\text{ is a factor of }f(x),\text{ therefore }f(-2)=0$
$\displaystyle (-2)^3+a(-2)+b=0$
$\displaystyle -8-2a+b=0\ \ \text{(i)}$
$\displaystyle \text{Also, }(x+3)\text{ is factor of }f(x),\text{ therefore }f(-3)=0$
$\displaystyle (-3)^3+a(-3)+b=0$
$\displaystyle -27-3a+b=0\ \ \text{(ii)}$
$\displaystyle \text{Subtract Eq.\ (ii) from Eq.\ (i), we get}$
$\displaystyle -8-2a+27+3a=0$
$\displaystyle a+19=0$
$\displaystyle a=-19$
$\displaystyle \text{Put the value of }a\text{ in Eq.\ (i), we get}$
$\displaystyle -8-2(-19)+b=0$
$\displaystyle -8+38+b=0$
$\displaystyle b=-30$

$\displaystyle \textbf{Question 15: }\text{Using the factor theorem, show that }(x - 2)\text{ is a factor of } \\ x^3 + x^2 - 4x - 4.\text{ Hence, factorise the polynomial completely.}\ \text{[ICSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=x^3+x^2-4x-4$
$\displaystyle \text{Using factor theorem, if }(x-2)\text{ is a factor of }f(x),\text{ then }f(2)=0$
$\displaystyle f(2)=(2)^3+(2)^2-4\times2-4$
$\displaystyle =8+4-8-4=0$
$\displaystyle \text{Hence, factor theorem is satisfied.}$
$\displaystyle \text{Now, using long division, divide }f(x)\text{ by }(x-2),\text{ we get}$
$\displaystyle \therefore f(x)=(x-2)(x^2+3x+2)$
$\displaystyle = (x-2)(x^2+2x+x+2)$
$\displaystyle = (x-2)[x(x+2)+1(x+2)]$
$\displaystyle = (x-2)(x+2)(x+1)$

$\displaystyle \textbf{Question 16: }\text{Use factor theorem to factorise }6x^3 + 17x^2 + 4x - 12\text{ completely.}\ \text{[ICSE 2020]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=6x^3+17x^2+4x-12$
$\displaystyle \text{Using Hit and Trial method, putting }x=-2,\text{ we get}$
$\displaystyle f(-2)=6(-2)^3+17(-2)^2+4(-2)-12$
$\displaystyle =-48+68-8-12=0$
$\displaystyle \text{So, by factor theorem, }(x+2)\text{ will be a factor of }f(x).$
$\displaystyle \text{Now, to find other factor, let us divide }f(x)\text{ by }(x+2)\text{ using long division method.}$
$\displaystyle \therefore f(x)=(x+2)(6x^2+5x-6)$
$\displaystyle \text{Now, factor of }6x^2+5x-6=6x^2+9x-4x-6$
$\displaystyle =3x(2x+3)-2(2x+3)$
$\displaystyle =(2x+3)(3x-2)$
$\displaystyle \text{Hence, }f(x)=(x+2)(2x+3)(3x-2).$

$\displaystyle \textbf{Question 17: }\text{If }x^2 - 4\text{ is a factor of polynomial }x^3 + x^2 - 4x - 4, \\ \text{ then its factors are}\ \text{[ICSE Semester I 2022]}$
$\displaystyle (a)\ (x - 2)(x + 2)(x + 1) \qquad (b)\ (x - 2)(x + 2)(x - 1)$
$\displaystyle (c)\ (x - 2)(x - 2)(x + 1) \qquad (d)\ (x - 2)(x - 2)(x - 1)$
$\displaystyle \text{Answer:}$
$\displaystyle (a)\ x^3+x^2-4x-4$
$\displaystyle \text{We can write, }x^2(x+1)-4(x+1)$
$\displaystyle =(x+1)(x^2-4)$
$\displaystyle =(x+1)(x-2)(x+2)$
$\displaystyle \text{Hence, }(x-2)(x+2)(x+1)\text{ is a factor of }x^3+x^2-4x-4.$

$\displaystyle \textbf{Question 18: }\text{If }(x + 2)\text{ is a factor of the polynomial }x^3 - kx^2 - 5x + 6,\text{ then the} \\ \text{value of }k\text{ is}\ \text{[ICSE Semester I 2022]}$
$\displaystyle (a)\ 1 \qquad (b)\ 2 \qquad (c)\ 3 \qquad (d)\ -2$
$\displaystyle \text{Answer:}$
$\displaystyle (b)\ \text{Let }f(x)=x^3-kx^2-5x+6$
$\displaystyle \text{Since, }(x+2)\text{ is a factor of }f(x).$
$\displaystyle f(-2)=0$
$\displaystyle (-2)^3-k(-2)^2-5(-2)+6=0$
$\displaystyle -8-4k+10+6=0$
$\displaystyle -4k+8=0$
$\displaystyle 4k=8$
$\displaystyle k=2$

$\displaystyle \textbf{Question 19: }\text{ }(x + 2)\text{ and }(x + 3)\text{ are two factors of the polynomial } \\ x^3 + 6x^2 + 11x + 6.\text{ If this polynomial is completely factorised, the result is} \\ \text{[ICSE Semester I 2022]}$
$\displaystyle (a)\ (x - 2)(x + 3)(x + 1) \qquad (b)\ (x + 2)(x - 3)(x - 1)$
$\displaystyle (c)\ (x + 2)(x + 3)(x - 1) \qquad (d)\ (x + 2)(x + 3)(x + 1)$
$\displaystyle \text{Answer:}$
$\displaystyle (d)\ \text{Let }f(x)=x^3+6x^2+11x+6$
$\displaystyle \text{Given, }(x+2)\text{ and }(x+3)\text{ are two factors of }f(x).$
$\displaystyle (x+2)(x+3)=x^2+3x+2x+6=x^2+5x+6$
$\displaystyle \therefore x^2+5x+6\text{ is also a factor of }f(x).$
$\displaystyle \text{Now, we divide }f(x)\text{ by }x^2+5x+6\text{ using long division method.}$
$\displaystyle \therefore f(x)=(x^2+5x+6)(x+1)$
$\displaystyle \text{Hence, }(x+2),(x+3)\text{ and }(x+1)\text{ completely factorise the polynomial }f(x).$

$\displaystyle \textbf{Question 20: }\text{If }x - 2\text{ is a factor of }x^3 - kx - 12,\text{ then the value of }k\text{ is}\ \text{[ICSE 2023]}$
$\displaystyle (a)\ 3 \qquad (b)\ 2 \qquad (c)\ -2 \qquad (d)\ -3$
$\displaystyle \text{Answer:}$
$\displaystyle (c)\ \text{Given, }(x-2)\text{ is a factor of }x^3-kx-12.$
$\displaystyle \Rightarrow x-2=0\Rightarrow x=2$
$\displaystyle \text{On substituting }x=2\text{ in given equation,}$
$\displaystyle 2^3-k\times2-12=0$
$\displaystyle 8-2k-12=0$
$\displaystyle -2k=4\Rightarrow k=-2$

$\displaystyle \textbf{Question 21: }\text{Find the value of }a\text{ if }x - a\text{ is a factor of the polynomial } \\ 3x^3 + x^2 - ax - 81.\ \text{[ICSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }(x-a)\text{ is a factor of the polynomial }3x^3+x^2-ax-81.$
$\displaystyle \text{Then, by factor theorem }x-a=0\Rightarrow x=a$
$\displaystyle \text{On substituting }x=a\text{ in given polynomial equation,}$
$\displaystyle 3a^3+a^2-a^2-81=0$
$\displaystyle 3a^3=81$
$\displaystyle a^3=27$
$\displaystyle a=3$

$\displaystyle \textbf{Question 22: }\text{Factorize completely using factor theorem } \\ 2x^3 - x^2 - 13x - 6.\ \text{[ICSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, expression is }2x^3-x^2-13x-6$
$\displaystyle \text{This expression can also be written as}$
$\displaystyle 2x^3+x^2-2x^2-x-12x-6$
$\displaystyle =x^2(2x+1)-x(2x+1)-6(2x+1)$
$\displaystyle =(2x+1)(x^2-x-6)$
$\displaystyle =(2x+1)(x^2-3x+2x-6)$
$\displaystyle =(2x+1)[x(x-3)+2(x-3)]$
$\displaystyle =(2x+1)(x-3)(x+2)$
$\displaystyle \text{Hence, the factors of }2x^3-x^2-13x-6\text{ are }(2x+1),(x-3)\text{ and }(x+2).$

$\displaystyle \textbf{Question 23: }\text{If }(x - 1)\text{ is a factor of }2x^2 - ax - 1,\text{ then the value of }a\text{ is}\ \text{[ICSE 2023]}$
$\displaystyle (a)\ -1 \qquad (b)\ 1 \qquad (c)\ 3 \qquad (d)\ -3$
$\displaystyle \text{Answer:}$
$\displaystyle (b)\ \text{Let }f(x)=2x^2-ax-1$
$\displaystyle \text{Since, }(x-1)\text{ is a factor of }f(x).$
$\displaystyle f(1)=0$
$\displaystyle 2(1)^2-a(1)-1=0$
$\displaystyle 2-a-1=0$
$\displaystyle a=1$
$\displaystyle \text{Hence, value of }a\text{ is }1.$

$\displaystyle \textbf{Question 24: }\text{What must be subtracted from the polynomial } \\ x^3 + x^2 - 2x + 1,\text{ so that the result is exactly divisible by }(x - 3)?\ \text{[ICSE 2024]}$
$\displaystyle (a)\ -31 \qquad (b)\ -30 \qquad (c)\ 30 \qquad (d)\ 31$
$\displaystyle \text{Answer:}$
$\displaystyle (d)\ \text{Given, }p(x)=x^3+x^2-2x+1$
$\displaystyle \text{By Remainder theorem}$
$\displaystyle \text{If }(x-3)\text{ is a factor of }p(x),\text{ then remainder will be }p(3).$
$\displaystyle p(3)=3^3+3^2-2\times3+1$
$\displaystyle =27+9-6+1=31$
$\displaystyle \therefore 31\text{ must be subtracted from }p(x),\text{ so that the result is divisible by }(x-3).$

$\displaystyle \textbf{Question 25: }\text{The polynomial }3x^3 + 8x^2 - 15x + k\text{ has }(x - 1)\text{ as a factor. Find the value of }k.\text{ Hence,} \\ \text{factorise the resulting polynomial completely.}\ \text{[ICSE 2024]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }p(x)=3x^3+8x^2-15x+k$
$\displaystyle \text{Since, }(x-1)\text{ is a factor of }p(x),\text{ therefore by factor theorem }p(1)=0$
$\displaystyle 3(1)^3+8(1)^2-15(1)+k=0$
$\displaystyle 3+8-15+k=0$
$\displaystyle k=4$
$\displaystyle \text{Given, }(x-1)\text{ is a factor of }p(x).$
$\displaystyle \therefore 3x^3+8x^2-15x+4$
$\displaystyle =(x-1)(3x^2+11x-4)$
$\displaystyle 3x^2+11x-4=3x^2+12x-x-4$
$\displaystyle =3x(x+4)-1(x+4)$
$\displaystyle =(x+4)(3x-1)$
$\displaystyle \text{Hence, }3x^3+8x^2-15x+4=(x-1)(x+4)(3x-1).$