Question 1: In each of the following cases, find the remainder when

i) $x^4-3x^2+2x+1$ is divided by $(x-1)$

ii) $x^3+3x^2-12x+4$ is divided by $(x-2)$

iii) $x^4+1$ is divided by $(x+1)$

i) Required Remainder = Value of given polynomial $x^4-3x^2+2x+1$ for $x = 1$

Therefore Remainder $= (1)^4-3(1)^2+2(1)+1 = 1$

ii) Required Remainder = Value of given polynomial $x^3+3x^2-12x+4$ for $x = 2$

Therefore Remainder $= (2)^3+3(2)^2-12(2)+4 = 0$

iii) Required Remainder = Value of given polynomial $x^4+1$ for $x = -1$

Therefore Remainder $= (-1)^4+1 = 2$

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Question 2: Show that

i) $(x-2)$ is a factor of $5x^2+15x-50$

ii) $(3x+2)$ is a factor of $3x^2-x-2$

i) If $(x-2)$ is a factor of $5x^2+15x-50$, then the remainder should be $0$ for $x = 2$

Remainder $= 5(2)^2+15(2)-50 = 50-50=0$

Hence $(x-2)$ is a factor of $5x^2+15x-50$

ii) If $(3x+2)$ is a factor of $3x^2-x-2$, then the remainder should be $0$ for $x = -\frac{2}{3}$

Remainder $= 3(-\frac{2}{3})^2-(-\frac{2}{3})-2 = \frac{4}{3}+\frac{2}{3}-2 = 0$

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

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Question 3: Find which of the following is a factor of $2x^3+3x^2-5x-6$

i) $(x+1)$   ii) $(2x-1)$   iii) $(x+2)$

i) If $(x+1)$ is a factor of $2x^3+3x^2-5x-6$, then the remainder should be $0$ for $x =-1$

Remainder = $2(-1)^3+3(-1)^2-5(-1)-6 = -2+3+5-6=0$

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

ii) If $(2x-1)$ is a factor of $2x^3+3x^2-5x-6$, then the remainder should be $0$ for $x =\frac{1}{2}$

Remainder = $2(\frac{1}{2})^3+3(\frac{1}{2})^2-5(\frac{1}{2})-6 = 1-\frac{3}{2}-6 \neq 0$

Hence $(2x-1)$ is NOT a factor of $2x^3+3x^2-5x-6$

iii) If $(x+2)$ is a factor of $2x^3+3x^2-5x-6$, then the remainder should be $0$ for $x =-2$

Remainder = $2(-2)^3+3(-2)^2-5(-2)-6 = -16+12+10-6= 0$

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

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Question 4: Find the value of $a \ or \ k$  if

i) $(2x+1)$ is a factor of $2x^2+ax-3$

ii) $(3x-4)$ is a factor of $3x^2+2x-k$

iii) $(2x+1)$ is a factor of $(3k+2)x^3+(k-1)$

iv) $(x-2)$ is a factor of $2x^5-6x^4-2ax^3+6ax^2+4ax+8$

i) $(2x+1)$ is a factor $\Rightarrow x = -\frac{1}{2}$

Therefore Remainder $=0$ for $x = -\frac{1}{2}$

$\Rightarrow 2( -\frac{1}{2})^2+a( -\frac{1}{2})-3 = 0$

$\Rightarrow \frac{1}{2}-\frac{a}{2}-3=0$

$\Rightarrow 1-a-6=0$

$\Rightarrow a = 5$

ii) $(3x-4)$ is a factor $\Rightarrow x = \frac{4}{3}$

Therefore Remainder $=0$ for $x = \frac{4}{3}$

$\Rightarrow 3(\frac{4}{3})^2+2(\frac{4}{3})-k=0$

$\Rightarrow \frac{16}{3}+\frac{8}{3}-k=0$

$\Rightarrow k = 8$

iii) If $(2x+1)$ is a factor $\Rightarrow x = -\frac{1}{2}$

Therefore Remainder $=0$ for $x = -\frac{1}{2}$

$\Rightarrow (3k+2)(-\frac{1}{2})^3+(k-1)=0$

$\Rightarrow -3k-2+8k-8=0$

$\Rightarrow k = 2$

iv) If $(x-2)$ is a factor $\Rightarrow x = 2$

Therefore Remainder $= 0$ for $x =2$

$\Rightarrow 2(2)^5-6(2)^4-2a(2)^3+6a(2)^2+4a(2)+8$

$\Rightarrow 64-96-16a+24a+8a+8 = 0$

$\Rightarrow a =\frac{24}{16} = \frac{3}{2}$

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Question 5: Find the value of $a \ and \ b$ , when

i) $(x-2)$ and $(x+3)$ are both factors of $x^3+ax^2+bx-12$

ii) $(x-1)$ and $(x+2)$ are both factors of $x^3+(3a+1)x^2+bx-18$

i) If $(x-2)$ is a factor

$\Rightarrow (2)^3+a(2)^2+b(2)-12=0$

$\Rightarrow 8+4a+2b-12=0$

$\Rightarrow 2a+b=2$ … … … … … i)

Similarly, if $(x+3)$ is a factor

$\Rightarrow (-3)^3+a(-3)^2+b(-3)-12=0$

$\Rightarrow -27+pa-3b-12=0$

$\Rightarrow 3a-b=13$  … … … … … ii)

Solving i) and ii)  we get $a = 3 \ and \ b =-4$

ii) If $(x-1)$ is a factor

$\Rightarrow (1)^3+(3a+1)(1)^2+b(1)-18=0$

$\Rightarrow 1 + 3a+1+b-18=0$

$\Rightarrow 3a+b=16$  … … … … … i)

If $(x+2)$ is a factor

$\Rightarrow (-2)^3+(3a+1)(-2)^2+b(-2)-18=k$

$\Rightarrow -8+12a+4-2b-18=0$

$\Rightarrow 6a-b=11$  … … … … … ii)

Solving i) and ii) $a = 3 \ and \ b = 7$

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Question 6:  When $x^3+2x^2-kx+4$ is divided by $(x-2)$ , the remainder is $k$ . Find $k$ .

When $x = 2,$  , Remainder $= 0$

$(2)^3+2(2)^2-k(2)+4 = k$

$\Rightarrow 20-2k = k$

$\Rightarrow k = \frac{20}{3}$

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Question 7: Find the value of $a$ , if the division of $ax^3+9x^2+4x-10$ by $(x+3)$ leaves a remainder of $5$ .

When $x = -3,$  , Remainder $= 5$

$a(-3)^3+9(-3)^2+4(-3)-10 = 5$

$\Rightarrow -27a+81-12-10=5$

$\Rightarrow 27a = 54$

$\Rightarrow a = 2$

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Question 8: 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 9: Find the value of $a \ and \ b$ , when $2x^3+ax^2+bx-2$ leaves a remainder $7 \ and \ 0$ when divided by $(2x-3)$ and $(x+2)$ respectively.

When $x=\frac{3}{2}$ , Remainder $= 7$

$\Rightarrow 2(\frac{3}{2})^3+a(\frac{3}{2})^2+b(\frac{3}{2})-2 = 7$

$\Rightarrow 27+9a+6b-8=28$

$\Rightarrow 3a+2b=3$ … … … … … i)

When $x = -2$ , Remainder $= 0$

$\Rightarrow 2(-2)^3+a(-2)^2+b(-2)-2= 0$

$\Rightarrow -16+4a-2b-2=0$

$\Rightarrow 2a-b=-9$  … … … … … ii)

Solving i) and ii) $a = 3 \ and \ b = -3$

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Question 10: What number should be added to $3x^3-5x^2+6x$ , so that when it is divided by $(x-3)$ , the remainder is $8$ .

Let $a$ be added to $3x^3-5x^2+6x$ , so that when it is divided by $(x-3)$ , the remainder is $8$

When $x =3$, Remainder is $8$

$\Rightarrow 3(3)^3-5(3)^2+6(3)+a=8$

$\Rightarrow 81-45+18+a=8$

$\Rightarrow a = -46$

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Question 11: What number should be subtracted to $x^3+3x^2-8x+14$ , so that when it is divided by $(x-2)$ , the remainder is $10$ .

Let $a$ be subtracted to $x^3+3x^2-8x+14$ , so that when it is divided by $(x-2)$ , the remainder is $10$

When $x =2$, Remainder is $10$

$\Rightarrow (2)^3+3(2)^2-8(2)+14-a=10$

$\Rightarrow 8+12-16+14-a=10$

$\Rightarrow a = 8$

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Question 12: The polynomials $2x^3-7x^2+ax-6$ and $x^3-8x^2+(2a+1)x-16$ leave the same remainder when divided by $(x-2)$ . Find the value of $a$ .

For polynomial $2x^3-7x^2+ax-6$:

When $x =2$, Remainder $= 2(2)^3-7(2)^2+a(2)-6 = 2a-18$

For polynomial $x^3-8x^2+(2a+1)x-16$:

When $x =2$, Remainder $= (2)^3-8(2)^2+(2a+1)(2)-16 = 4a-38$

Therefore $2a-18=4a-38 \Rightarrow a = 10$

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Question 13: 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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