Class 10: Remainder and Factor Theorems – ICSE Board Problems – ICSE / ISC / CBSE Mathematics Portal for K12 Students

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]

Answer:

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]

Answer:

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]

Answer:

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]

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

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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]