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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)

Answer:

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

Answer:

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)

Answer:

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

Answer:

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

Answer:

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 .

Answer:

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 .

Answer:

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]

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 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.

Answer:

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 .

Answer:

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 .

Answer:

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 .

Answer:

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]

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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