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]

Answer:

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

Answer:

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]

Answer:

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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