Question 1: 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 $a \ and \ b$ .     [2005]

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) $a = -3 \ and \ b = -1$

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Question 2: 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 $a \ and \ b$.     [2013]

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 $a = 5 \ and \ b = -11$

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Question 3: Find the value of  $a$ , if  $(x-a)$ is a factor of  $x^3-ax^2+x+2$.     [2003]

When $x = a$, Remainder  $= 0$

Therefore  $(a)^3-a(a)^2+(a)+2 =0$

$\Rightarrow a^3-a^3+a+2=0$

$\Rightarrow a = 2$

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Question 4: 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 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 $p$ .     [2010]

When $x=3$

$\text{Remainder}_1 = (3)^3-p(3)^2+(3)+6$

$= 27-9p+9$

$= 36-9p$

$\text{Remainder}_2 = 2(3)^3-(3)^2-(p+3)(3)-6$

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

$=30-39$

$\text{Given Remainder}_1 = \text{Remainder}_1$

$36-9p=30-3p$

$6=6p \Rightarrow p =1$

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Question 6: 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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Question 7: 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.    [2011]

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 $x^3+2x^2-kx+10$

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