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

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 $x^3-7x^2+14x-8 =(x-1)(x-2)(x-4)$

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

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 $x^3+10x^2-37x+26 =(x-1)(x-2)(x+13)$

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Question 3: Find the value of $m$ when $x^3+3x^2-mx+4$ is divided by $(x-2)$ , the remainder is $(m+3)$ .

When $x = 2$

Remainder $\Rightarrow (2)^3+3(2)^2-m(2)+4 = m+3$

$8+12-2m+4=m+3$

$\Rightarrow m = 7$

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Question 4: What should be subtracted from $3x^3-8x^2+4x-3$ , so that the resulting expression has $(x+2)$ as a factor.

Let $a$ be subtracted

When $x = -2$

Remainder $\Rightarrow 3(-2)^3-8(-2)^2+4(-2)-3-a =0$

$\Rightarrow -24-32-8-3=a$

$\Rightarrow a= - 67$

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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.

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 $x^3+2x^2-5x-6 =(x-1)(x+3)(x-2)$

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Question 6: If $(x-2)$ is a factor of $x^2+ax+b$ and $a+b=1$ , find the values of $a \ and \ b$ .

When $x = 2$

Remainder: $(2)^2+a(2)+b =0$

$4+2a+b=0$

Given $a+b=1$

Therefore $4+a+(a+b)=0$

$\Rightarrow a = -5$

Solving for $b = 6$

Hence $a = -5 \ and \ b = 6$

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

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 $x^3+6x^2+11x+6 =(x+1)(x+2)(x+3)$

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Question 8: Find the value of $m$ , if $mx^3+2x^2-3$ and $x^2-mx+4$ leave the same remainder when each is divided by $(x-2)$ .

When $x = 2$

Remainder 1 $= m(2)^3+2(2)^2-3 = 8m+5$

Remainder 2 $= (2)^2-m(2)+4 = 8-m$

Given Remainder 1 = Remainder 2

$8m+5=8-m \Rightarrow m = \frac{1}{3}$

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Question 9: The polynomial $px^3+4x^2-3x+q$ is completely divisible by $(x^2-1)$ . Find the value of $p \ and \ q$ . Also for these values of $p \ and \ q$ , factorize the given polynomial completely.

$(x^2-1)$ is a factor of $px^3+4x^2-3x+q$

$\Rightarrow (x-1)(x+1)$ is a factor of $px^3+4x^2-3x+q$

$\Rightarrow (x-1) \ and \ (x+1)$ are factors of  $px^3+4x^2-3x+q$

When $x = 1$

$p(1)^3+4(1)^2-3(1)+q = 0$

$\Rightarrow p+q=-1$ … … … … … i)

When $x = -1$

$p(-1)^3+4(-1)^2-3(-1)+q = 0$

$\Rightarrow p-q=7$… … … … … ii)

Solving i) and ii) we get $p = 3 \ and \ q = -4$

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Question 10: Find the number which should be added to $x^2+x+3$ so that the resulting polynomials completely divisible by $(x+3)$ .

Let $a$ be added to the polynomial.

When $x = -3$

Remainder: $(-3)^2+(-3)+3+a = 0 \Rightarrow a = -9$

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Question 11: When the polynomial $x^3+2x^2-5ax-7$ is divided by $(x-1)$ , the remainder is $A$ . When the polynomial $x^3+ax^2-12x+16$ is divided by $(x+2)$ , the remainder is $B$ . Find the value of $a$  is $2A+B=0$ .

When $x = 1$

Remainder: $(1)^3+2(1)^2-5a(1)-7=A$

$\Rightarrow 1+2-5a-7=A$

$\Rightarrow -5a-4=A$ … … … … … i)

When $x = -2$

Remainder: $(-2)^3+a(-2)^2-12(-2)+16=B$

$\Rightarrow -8+4a+24+16=B$

$\Rightarrow 4a+32=B$ … … … … … ii)

Given $2A+B = 0$

$\Rightarrow 2(-5a-4)+(4a+32)=0$

$\Rightarrow -10a-8+4a+32=0$

$\Rightarrow a = 4$

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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.

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

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

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

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

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

$40a = 160$

$\Rightarrow a = 4$

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

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Question 13: 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 $p$ .     [2010]

When $x=3$

Remainder1 $= (3)^3-p(3)^2+(3)+6$

$= 27-9p+9$

$= 36-9p$

Remainder2 $= 2(3)^3-(3)^2-(p+3)(3)-6$

$=54-9-3p-9-6$

$=30-39$

Given Remainder1 = Remainder 2

$36-9p=30-3p$

$6=6p \Rightarrow p =1$

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Question 14: Use the remainder theorem to factorize the following expression: $2x^3+x^2-13x+6$ .     [2010]

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 $2x^3+x^2-13x+6 = (x-2)(x+3)(2x-1)$

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