Class 10: Remainder and Factor Theorem

Question 1: Show $(x-1)$ is a factor of $x^3-7x^2+14x-8$ . Factorize the polynomial.

Answer:

For $x = 1$ ,

Remainder: $= (1)^3-7(1)^2+14(1)-8 = 1-7+14-8=0$

Hence $(x-1)$ is a factor of $x^3-7x^2+14x-8$

$x-1 ) \overline {x^3-7x^2+14x-8} (x^2-6x+8$
$(-) \ \ \underline {x^3-x^2}$
$-6x^2+14x-8$
$(-) \ \ \underline{-6x^2+6x}$
$8x-8$
$(-) \ \ \underline{ 8x-8}$
$\times$

$x^3-7x^2+14x-8 = (x-1)(x^2-6x+8)$

$= (x-1)(x^2-2x-4x+8)$

$= (x-1)[x(x-2)-4(x-2)]$

$= (x-1)(x-2)(x-4)$

Hence $x^3-7x^2+14x-8 =(x-1)(x-2)(x-4)$

$\\$

Question 2: Using remainder theorem, factorize $x^3+10x^2-37x+26$ completely. [2014]

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

$\\$

$\displaystyle \textbf{Question 3: } \text{Find the value of }m\text{ when }x^3+3x^2-mx+4\text{ is divided by }(x-2),$
$\displaystyle \text{the remainder is }(m+3).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }f(x)=x^3+3x^2-mx+4$
$\displaystyle \text{When }x=2,\text{ remainder}=m+3$
$\displaystyle f(2)=m+3$
$\displaystyle (2)^3+3(2)^2-m(2)+4=m+3$
$\displaystyle 8+12-2m+4=m+3$
$\displaystyle 24-2m=m+3$
$\displaystyle 21=3m$
$\displaystyle m=7$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: } \text{What should be subtracted from }3x^3-8x^2+4x-3,\text{ so that}$
$\displaystyle \text{the resulting expression has }(x+2)\text{ as a factor?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }a\text{ be subtracted from the polynomial.}$
$\displaystyle \text{Since }(x+2)\text{ is a factor, put }x=-2$
$\displaystyle 3(-2)^3-8(-2)^2+4(-2)-3-a=0$
$\displaystyle -24-32-8-3-a=0$
$\displaystyle -67-a=0$
$\displaystyle a=-67$
$\displaystyle \therefore \text{The required number is }-67$
$\displaystyle \\$

Question 5: If $(x+1)$ and $(x-2)$ are factors of $x^3+(a+1)x^2-(b-2)x-6$ , find the values of $a \ and \ b$ . Factorize the polynomial also.

Answer:

When $x = -1$

Remainder: $(-1)^3+(a+1)(-1)^2-(b-2)(-1)-6 =0$

$\Rightarrow -1+(a+1)+(b-2)-6=0$

$\Rightarrow a+b=8$ … … … … … … i)

When $x = 2$

Remainder: $(2)^3+(a+1)(2)^2-(b-2)(2)-6 =0$

$\Rightarrow 8+4a+4-2b+4-6=0$

$\Rightarrow 2a-b=-5$ … … … … … … ii)

Solving i) and ii) we get $a = 1 \ and \ b = 7$

Substituting the values in the polynomial we get $x^3+2x^2-5x-6$

Given $(x+1)$ is a factor of $x^3+2x^2-5x-6$

$x+1 ) \overline {x^3+2x^2-5x-6} (x^2+x-6$
$(-) \ \ \underline {x^3+x^2}$
$x^2-5x-6$
$(-) \ \ \underline{x^2+x}$
$-6x-6$
$(-) \ \ \underline{ -6x-6}$
$\times$

$x^3+2x^2-5x-6 = (x+1)(x^2+x-6)$

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

$= (x+1)[x(x+3)-2(x+3)]$

$= (x+1)(x+3)(x-2)$

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

$\\$

$\displaystyle \textbf{Question 6: } \text{If }(x-2)\text{ is a factor of }x^2+ax+b\text{ and }a+b=1,$
$\displaystyle \text{find the values of }a\text{ and }b.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }(x-2)\text{ is a factor, put }x=2$
$\displaystyle (2)^2+a(2)+b=0$
$\displaystyle 4+2a+b=0\qquad\ldots\ldots\ldots\text{(1)}$
$\displaystyle \text{Given }a+b=1\qquad\ldots\ldots\ldots\text{(2)}$
$\displaystyle \text{Substituting (2) into (1),}$
$\displaystyle 4+a+(a+b)=0$
$\displaystyle 4+a+1=0$
$\displaystyle a=-5$
$\displaystyle \text{Putting }a=-5\text{ in (2),}$
$\displaystyle -5+b=1$
$\displaystyle b=6$
$\displaystyle \therefore a=-5\text{ and }b=6$
$\displaystyle \\$

Question 7: Using remainder theorem, factorize $x^3+6x^2+11x+6$ completely.

Answer:

For $x = -1$ ,

Remainder: $= (-1)^3+6(-1)^2+11(-1)+6 = -1+6-11+6=0$

Hence $(x+1)$ is a factor of $x^3+6x^2+11x+6$

$x+1 ) \overline {x^3+6x^2+11x+6} (x^2+5x+6$
$(-) \ \ \underline {x^3+x^2}$
$5x^2+11x+6$
$(-) \ \ \underline{5x^2+5x}$
$6x + 6$
$(-) \ \ \underline{ 6x+6}$
$\times$

$x^3+6x^2+11x+6 = (x+1)(x^2+5x+6)$

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

$= (x+1)[x(x+2)+3(x+2)]$

$= (x+1)(x+2)(x+3)$

Hence $x^3+6x^2+11x+6 =(x+1)(x+2)(x+3)$

$\\$

$\displaystyle \textbf{Question 8: } \text{Find the value of }m,\text{ if }mx^3+2x^2-3\text{ and }x^2-mx+4$
$\displaystyle \text{leave the same remainder when each is divided by }(x-2).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=2,$
$\displaystyle \text{Remainder}_1=m(2)^3+2(2)^2-3=8m+8-3=8m+5$
$\displaystyle \text{Remainder}_2=(2)^2-m(2)+4=4-2m+4=8-2m$
$\displaystyle \text{Since the remainders are equal,}$
$\displaystyle 8m+5=8-2m$
$\displaystyle 10m=3$
$\displaystyle m=\frac{3}{10}$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: } \text{The polynomial }px^3+4x^2-3x+q\text{ is completely divisible by }$
$\displaystyle (x^2-1). \ \text{Find }p\text{ and }q.\text{ Also for these values of }p\text{ and }q,\text{ factorise it completely.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }x^2-1=(x-1)(x+1),\text{ both }(x-1)\text{ and }(x+1)\text{ are factors.}$
$\displaystyle \text{Let }f(x)=px^3+4x^2-3x+q$
$\displaystyle \text{Since }(x-1)\text{ is a factor, }f(1)=0$
$\displaystyle p(1)^3+4(1)^2-3(1)+q=0$
$\displaystyle p+4-3+q=0$
$\displaystyle p+q=-1\qquad\ldots\ldots\ldots\text{(1)}$
$\displaystyle \text{Since }(x+1)\text{ is a factor, }f(-1)=0$
$\displaystyle p(-1)^3+4(-1)^2-3(-1)+q=0$
$\displaystyle -p+4+3+q=0$
$\displaystyle -p+q=-7\qquad\ldots\ldots\ldots\text{(2)}$
$\displaystyle \text{Solving (1) and (2), we get}$
$\displaystyle p=3\text{ and }q=-4$
$\displaystyle \therefore f(x)=3x^3+4x^2-3x-4$
$\displaystyle 3x^3+4x^2-3x-4=x^2(3x+4)-1(3x+4)$
$\displaystyle =(x^2-1)(3x+4)$
$\displaystyle =(x-1)(x+1)(3x+4)$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: } \text{Find the number which should be added to }x^2+x+3\text{ so that}$
$\displaystyle \text{the resulting polynomial is completely divisible by }(x+3).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }a\text{ be added to the polynomial.}$
$\displaystyle \text{Since the resulting polynomial is divisible by }(x+3),\text{ put }x=-3$
$\displaystyle (-3)^2+(-3)+3+a=0$
$\displaystyle 9-3+3+a=0$
$\displaystyle 9+a=0$
$\displaystyle a=-9$
$\displaystyle \therefore \text{The required number is }-9$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: } \text{When }x^3+2x^2-5ax-7\text{ is divided by }(x-1),\text{ the remainder is }A.$
$\displaystyle \text{When }x^3+ax^2-12x+16\text{ is divided by }(x+2),\text{ the remainder is }B.$
$\displaystyle \text{Find }a\text{ if }2A+B=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For }x^3+2x^2-5ax-7,\text{ put }x=1$
$\displaystyle A=(1)^3+2(1)^2-5a(1)-7$
$\displaystyle A=1+2-5a-7$
$\displaystyle A=-5a-4\qquad\ldots\ldots\ldots\text{(1)}$
$\displaystyle \text{For }x^3+ax^2-12x+16,\text{ put }x=-2$
$\displaystyle B=(-2)^3+a(-2)^2-12(-2)+16$
$\displaystyle B=-8+4a+24+16$
$\displaystyle B=4a+32\qquad\ldots\ldots\ldots\text{(2)}$
$\displaystyle \text{Given, }2A+B=0$
$\displaystyle 2(-5a-4)+(4a+32)=0$
$\displaystyle -10a-8+4a+32=0$
$\displaystyle -6a+24=0$
$\displaystyle a=4$
$\displaystyle \\$

Question 12: $(3x+5)$ is the factor of the polynomial $(a-1)x^3+(a+1)x^2-(2a+1)x-15$ . Find the value of $a$ and factorize the give polynomial.

Answer:

When $x =-\frac{5}{3}$

Remainder: $\displaystyle (a-1)(-\frac{5}{3})^3+(a+1)(-\frac{5}{3})^2-(2a+1)(-\frac{5}{3})-15=0$

$\displaystyle -\frac{125}{27}(a-1)+\frac{25}{9}(a+1)+\frac{5}{3}(2a+1)-15=0$

$-125(a-1)+75(a+1)+45(2a+1)-405=0$

$a(-125+75+90)+125+75+45-405=0$

$40a = 160$

$\Rightarrow a = 4$

Hence $(3x+5)$ is a factor of $3x^3+5x^2-9x-15$

$3x+5 ) \overline {3x^3+5x^2-9x-15} (x^2-3$
$(-) \ \ \underline {3x^3+5x^2}$
$-9x-15$
$(-) \ \ \underline{-9x-15}$
$\times$

$3x^3+5x^2-9x-15 = (3x+5)(x^2-3)$

$= (3x+5)(x+\sqrt{3})(x-\sqrt{3})$

Hence $3x^3+5x^2-9x-15 =(3x+5)(x+\sqrt{3})(x-\sqrt{3})$

$\\$

$\displaystyle \textbf{Question 13: } \text{When divided by }(x-3),\text{ the polynomials }x^3-px^2+x+6$
$\displaystyle \text{and }2x^3-x^2-(p+3)x-6\text{ leave the same remainder. Find }p.\hfill\text{[ICSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{When }x=3,$
$\displaystyle \text{Remainder}_1=(3)^3-p(3)^2+(3)+6$
$\displaystyle =27-9p+3+6$
$\displaystyle =36-9p$
$\displaystyle \text{Remainder}_2=2(3)^3-(3)^2-(p+3)(3)-6$
$\displaystyle =54-9-3p-9-6$
$\displaystyle =30-3p$
$\displaystyle \text{Since the remainders are equal,}$
$\displaystyle 36-9p=30-3p$
$\displaystyle 6=6p$
$\displaystyle p=1$
$\displaystyle \\$