Class 10: Remainder and Factor Theorems

$\displaystyle \textbf{Question 1: } \text{Find the remainder when }x^2-8x+4\text{ is divided by }(2x+1).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Step 1: }2x+1=0\Rightarrow x=-\frac{1}{2}$
$\displaystyle \text{Step 2: Required remainder = Value of }x^2-8x+4\text{ at }x=-\frac{1}{2}$
$\displaystyle \text{Remainder}=\left(-\frac{1}{2}\right)^2-8\left(-\frac{1}{2}\right)+4$
$\displaystyle =\frac{1}{4}+4+4=\frac{33}{4}=8\frac{1}{4}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: } \text{Find the value of }a\text{ if the division of }ax^3+9x^2+4x-10$
$\displaystyle \text{by }(x+3)\text{ leaves a remainder of }5.$
$\displaystyle \text{Answer:}$
$\displaystyle x+3=0\Rightarrow x=-3$
$\displaystyle \text{Since the remainder is }5,\text{ the value of the polynomial at }x=-3\text{ is }5$
$\displaystyle a(-3)^3+9(-3)^2+4(-3)-10=5$
$\displaystyle -27a+81-12-10=5$
$\displaystyle -27a=-54$
$\displaystyle a=2$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: } \text{When the polynomial }2x^3-kx^2+(5k-3)x-8\text{ is divided by }$
$\displaystyle (x-2), \text{the remainder is }14.\text{ Find the value of }k.$
$\displaystyle \text{Answer:}$
$\displaystyle x-2=0\Rightarrow x=2$
$\displaystyle \text{Since the remainder is }14,\text{ the value of the polynomial at }x=2\text{ is }14$
$\displaystyle 2(2)^3-k(2)^2+(5k-3)(2)-8=14$
$\displaystyle 16-4k+10k-6-8=14$
$\displaystyle 6k+2=14$
$\displaystyle 6k=12$
$\displaystyle k=2$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: } \text{The polynomials }3x^3-ax^2+5x-13\text{ and }(a+1)x^2-7x+5$
$\displaystyle \text{leave the same remainder when divided by }(x-3).\text{ Find the value of }a.$
$\displaystyle \text{Answer:}$
$\displaystyle x-3=0\Rightarrow x=3$
$\displaystyle \text{Since the remainders are same,}$
$\displaystyle 3(3)^3-a(3)^2+5(3)-13=(a+1)(3)^2-7(3)+5$
$\displaystyle 81-9a+15-13=9a+9-21+5$
$\displaystyle 83-9a=9a-7$
$\displaystyle 18a=90$
$\displaystyle a=5$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: } \text{When }f(x)=x^3+ax^2-bx-8\text{ is divided by }(x-2),$
$\displaystyle \text{the remainder is zero and when divided by }(x+1),\text{ the remainder is }-30.$
$\displaystyle \text{Find the values of }a\text{ and }b.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }x-2=0\Rightarrow x=2$
$\displaystyle f(2)=0$
$\displaystyle (2)^3+a(2)^2-b(2)-8=0$
$\displaystyle 8+4a-2b-8=0$
$\displaystyle 4a-2b=0$
$\displaystyle 2a-b=0\qquad\ldots\ldots\ldots\text{(1)}$
$\displaystyle \text{Since }x+1=0\Rightarrow x=-1$
$\displaystyle f(-1)=-30$
$\displaystyle (-1)^3+a(-1)^2-b(-1)-8=-30$
$\displaystyle -1+a+b-8=-30$
$\displaystyle a+b=-21\qquad\ldots\ldots\ldots\text{(2)}$
$\displaystyle \text{Solving (1) and (2), we get}$
$\displaystyle a=-7\text{ and }b=-14$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: } \text{What number should be added to }2x^3-3x^2+x\text{ so that when}$
$\displaystyle \text{the resulting polynomial is divided by }(x-2),\text{ the remainder is }3?$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the number added be }k$
$\displaystyle \text{Then the resulting polynomial is }2x^3-3x^2+x+k$
$\displaystyle \text{Since the remainder is }3\text{ when divided by }(x-2),\text{ put }x=2$
$\displaystyle 2(2)^3-3(2)^2+2+k=3$
$\displaystyle 16-12+2+k=3$
$\displaystyle 6+k=3$
$\displaystyle k=-3$
$\displaystyle \therefore \text{The required number is }-3$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: } \text{Determine whether }(x-1)\text{ is a factor of } \\ x^6-x^5+x^4+x^3-x^2-x+1\text{ or not.}$
$\displaystyle \text{Answer:}$
$\displaystyle x-1=0\Rightarrow x=1$
$\displaystyle \text{Remainder}=f(1)$
$\displaystyle =(1)^6-(1)^5+(1)^4+(1)^3-(1)^2-(1)+1$
$\displaystyle =1-1+1+1-1-1+1=1$
$\displaystyle \text{Since the remainder }\neq0,\ (x-1)\text{ is not a factor of the given polynomial.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: } \text{If }(x-2)\text{ is a factor of }x^2-7x+2a,\text{ find the value of }a.$
$\displaystyle \text{Answer:}$
$\displaystyle x-2=0\Rightarrow x=2$
$\displaystyle \text{Since }(x-2)\text{ is a factor, the remainder is }0$
$\displaystyle (2)^2-7(2)+2a=0$
$\displaystyle 4-14+2a=0$
$\displaystyle 2a=10$
$\displaystyle a=5$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: } \text{Find the value of }k\text{ if }(x-2)\text{ is a factor of }x^3+2x^2-kx+10.$
$\displaystyle \text{Hence, determine whether }(x+5)\text{ is also a factor.}\hfill\text{[ICSE 2011]}$
$\displaystyle \text{Answer:}$
$\displaystyle x-2=0\Rightarrow x=2$
$\displaystyle \text{Since }(x-2)\text{ is a factor, the remainder is }0$
$\displaystyle (2)^3+2(2)^2-k(2)+10=0$
$\displaystyle 8+8-2k+10=0$
$\displaystyle 26-2k=0$
$\displaystyle k=13$
$\displaystyle \text{Substituting }k=13,\text{ the polynomial becomes }x^3+2x^2-13x+10$
$\displaystyle \text{To check whether }(x+5)\text{ is a factor, put }x=-5$
$\displaystyle (-5)^3+2(-5)^2-13(-5)+10$
$\displaystyle =-125+50+65+10=0$
$\displaystyle \text{Since the remainder is }0,\ (x+5)\text{ is also a factor.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: } \text{Given that }(x+2)\text{ and }(x-3)\text{ are factors of }x^3+ax+b,$
$\displaystyle \text{calculate the values of }a\text{ and }b.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }(x+2)\text{ is a factor, put }x=-2$
$\displaystyle (-2)^3+a(-2)+b=0$
$\displaystyle -8-2a+b=0$
$\displaystyle -2a+b=8\qquad\ldots\ldots\ldots\text{(1)}$
$\displaystyle \text{Since }(x-3)\text{ is a factor, put }x=3$
$\displaystyle (3)^3+a(3)+b=0$
$\displaystyle 27+3a+b=0$
$\displaystyle 3a+b=-27\qquad\ldots\ldots\ldots\text{(2)}$
$\displaystyle \text{Solving (1) and (2), we get}$
$\displaystyle a=-7\text{ and }b=-6$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: } \text{Polynomial }x^3-ax^2+bx-6\text{ leaves remainder }-8\text{ when divided by }$
$\displaystyle (x-1) \ \text{and }(x-2)\text{ is a factor of it. Find the values of }a\text{ and }b.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When divided by }(x-1),\text{ put }x=1$
$\displaystyle (1)^3-a(1)^2+b(1)-6=-8$
$\displaystyle 1-a+b-6=-8$
$\displaystyle -a+b=-3$
$\displaystyle a-b=3\qquad\ldots\ldots\ldots\text{(1)}$
$\displaystyle \text{Since }(x-2)\text{ is a factor, put }x=2$
$\displaystyle (2)^3-a(2)^2+b(2)-6=0$
$\displaystyle 8-4a+2b-6=0$
$\displaystyle 2a-b=1\qquad\ldots\ldots\ldots\text{(2)}$
$\displaystyle \text{Subtracting (1) from (2),}$
$\displaystyle a=-2$
$\displaystyle \text{Putting }a=-2\text{ in (1),}$
$\displaystyle -2-b=3$
$\displaystyle b=-5$
$\displaystyle \therefore a=-2\text{ and }b=-5$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: } \text{Using the Factor Theorem, show that }(x-2)\text{ is a factor of }$
$\displaystyle 3x^2-5x-2. \ \text{Hence, factorize the given expression.}$
$\displaystyle \text{Answer:}$
$\displaystyle x-2=0\Rightarrow x=2$
$\displaystyle \text{Remainder}=3(2)^2-5(2)-2$
$\displaystyle =12-10-2=0$
$\displaystyle \therefore (x-2)\text{ is a factor of }3x^2-5x-2$
$\displaystyle \text{Now, }3x^2-5x-2=3x^2-6x+x-2$
$\displaystyle =3x(x-2)+1(x-2)$
$\displaystyle =(x-2)(3x+1)$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: } \text{Show that }(2x+7)\text{ is a factor of }2x^3+5x^2-11x-14.$
$\displaystyle \text{Hence factorize the given expression completely, using the Factor Theorem.}\hfill\text{[ICSE 2006]}$
$\displaystyle \text{Answer:}$
$\displaystyle 2x+7=0\Rightarrow x=-\frac{7}{2}$
$\displaystyle \text{Remainder}=2\left(-\frac{7}{2}\right)^3+5\left(-\frac{7}{2}\right)^2-11\left(-\frac{7}{2}\right)-14$
$\displaystyle =-\frac{343}{4}+\frac{245}{4}+\frac{77}{2}-14$
$\displaystyle =\frac{-343+245+154-56}{4}=0$
$\displaystyle \therefore (2x+7)\text{ is a factor of }2x^3+5x^2-11x-14$
$\displaystyle 2x^3+5x^2-11x-14=(2x+7)(x^2-x-2)$
$\displaystyle =(2x+7)(x^2-2x+x-2)$
$\displaystyle =(2x+7)\{x(x-2)+1(x-2)\}$
$\displaystyle =(2x+7)(x-2)(x+1)$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: } \text{Using the Remainder Theorem, factorize the expression }$
$\displaystyle 2x^3+x^2-2x-1 \ \text{completely.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For }x=1,\text{ the value of the given expression is}$
$\displaystyle 2(1)^3+(1)^2-2(1)-1=2+1-2-1=0$
$\displaystyle \therefore (x-1)\text{ is a factor of }2x^3+x^2-2x-1$
$\displaystyle 2x^3+x^2-2x-1=(x-1)(2x^2+3x+1)$
$\displaystyle =(x-1)(2x^2+2x+x+1)$
$\displaystyle =(x-1)\{2x(x+1)+1(x+1)\}$
$\displaystyle =(x-1)(x+1)(2x+1)$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: } \text{Find the values of }a\text{ and }b\text{ so that }x^3+ax^2+bx-45$
$\displaystyle \text{has }(x-1)\text{ and }(x+5)\text{ as its factors. Hence, factorize it completely.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }(x-1)\text{ is a factor, put }x=1$
$\displaystyle (1)^3+a(1)^2+b(1)-45=0$
$\displaystyle a+b=44\qquad\ldots\ldots\ldots\text{(1)}$
$\displaystyle \text{Since }(x+5)\text{ is a factor, put }x=-5$
$\displaystyle (-5)^3+a(-5)^2+b(-5)-45=0$
$\displaystyle -125+25a-5b-45=0$
$\displaystyle 5a-b=34\qquad\ldots\ldots\ldots\text{(2)}$
$\displaystyle \text{Solving (1) and (2), we get}$
$\displaystyle a=13\text{ and }b=31$
$\displaystyle \therefore x^3+ax^2+bx-45=x^3+13x^2+31x-45$
$\displaystyle x^3+13x^2+31x-45=(x-1)(x^2+14x+45)$
$\displaystyle =(x-1)(x^2+9x+5x+45)$
$\displaystyle =(x-1)\{x(x+9)+5(x+9)\}$
$\displaystyle =(x-1)(x+9)(x+5)$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: } \text{If }(x-2)\text{ is a factor of }2x^3-x^2-px-2,$
$\displaystyle \text{(i) find the value of }p.\text{ (ii) With this value of }p,\text{ factorize it completely.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Since }(x-2)\text{ is a factor, put }x=2$
$\displaystyle 2(2)^3-(2)^2-p(2)-2=0$
$\displaystyle 16-4-2p-2=0$
$\displaystyle 10-2p=0$
$\displaystyle p=5$
$\displaystyle \text{(ii) Substituting }p=5,\text{ the polynomial becomes }2x^3-x^2-5x-2$
$\displaystyle 2x^3-x^2-5x-2=(x-2)(2x^2+3x+1)$
$\displaystyle =(x-2)(2x^2+2x+x+1)$
$\displaystyle =(x-2)\{2x(x+1)+1(x+1)\}$
$\displaystyle =(x-2)(x+1)(2x+1)$
$\displaystyle \\$