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Question 1: Show (x-1) is a factor of x^3-7x^2+14x-8 . Factorize the polynomial.

Answer:

For x = 1 ,

Remainder: = (1)^3-7(1)^2+14(1)-8 = 1-7+14-8=0 

Hence (x-1)  is a factor of  x^3-7x^2+14x-8

  • x-1 ) \overline {x^3-7x^2+14x-8} (x^2-6x+8
  •  (-) \ \  \underline {x^3-x^2}  
  •                   -6x^2+14x-8
  •          (-) \ \   \underline{-6x^2+6x}
  •                              8x-8
  •                      (-) \ \   \underline{ 8x-8}
  •                                      \times

x^3-7x^2+14x-8 = (x-1)(x^2-6x+8) 

 = (x-1)(x^2-2x-4x+8) 

 = (x-1)[x(x-2)-4(x-2)] 

 = (x-1)(x-2)(x-4) 

Hence x^3-7x^2+14x-8 =(x-1)(x-2)(x-4) 

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Question 2: 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 3: Find the value of m when x^3+3x^2-mx+4 is divided by  (x-2) , the remainder is (m+3) .11

Answer:

When x = 2

Remainder \Rightarrow (2)^3+3(2)^2-m(2)+4 = m+3

8+12-2m+4=m+3

\Rightarrow m = 7

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Question 4: What should be subtracted from 3x^3-8x^2+4x-3 , so that the resulting expression has (x+2) as a factor.

Answer:

Let a  be subtracted

When x = -2

Remainder \Rightarrow 3(-2)^3-8(-2)^2+4(-2)-3-a =0

\Rightarrow -24-32-8-3=a

\Rightarrow  a= - 67

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Question 5: If (x+1) and (x-2) are factors of x^3+(a+1)x^2-(b-2)x-6 , find the values of a \ and \  b . Factorize the polynomial also.

Answer:

When x = -1

Remainder: (-1)^3+(a+1)(-1)^2-(b-2)(-1)-6 =0

\Rightarrow -1+(a+1)+(b-2)-6=0

\Rightarrow a+b=8 … … … … … … i)

When x = 2

Remainder: (2)^3+(a+1)(2)^2-(b-2)(2)-6 =0

\Rightarrow 8+4a+4-2b+4-6=0

\Rightarrow 2a-b=-5 … … … … … … ii)

Solving i) and ii) we get a = 1 \ and \ b = 7

Substituting the values in the polynomial we get x^3+2x^2-5x-6

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

  • x+1 ) \overline {x^3+2x^2-5x-6} (x^2+x-6
  •  (-) \ \  \underline {x^3+x^2}  
  •                   x^2-5x-6
  •          (-) \ \   \underline{x^2+x}
  •                              -6x-6
  •                      (-) \ \   \underline{ -6x-6}
  •                                      \times

x^3+2x^2-5x-6 = (x+1)(x^2+x-6) 

 = (x+1)(x^2+3x-2x-6) 

 = (x+1)[x(x+3)-2(x+3)] 

 = (x+1)(x+3)(x-2) 

Hence x^3+2x^2-5x-6 =(x-1)(x+3)(x-2)

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Question 6: If (x-2) is a factor of x^2+ax+b and a+b=1 , find the values of a \ and \ b .

Answer:

When x = 2

Remainder: (2)^2+a(2)+b =0

4+2a+b=0 41

Given a+b=1

Therefore 4+a+(a+b)=0

\Rightarrow a = -5

Solving for b = 6

Hence a = -5 \ and \ b = 6

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Question 7: Using remainder theorem, factorize x^3+6x^2+11x+6 completely. 

Answer:

For x = -1 ,

Remainder: = (-1)^3+6(-1)^2+11(-1)+6 = -1+6-11+6=0 

Hence (x+1)  is a factor of  x^3+6x^2+11x+6

  • x+1 ) \overline {x^3+6x^2+11x+6} (x^2+5x+6
  •  (-) \ \  \underline {x^3+x^2}  
  •                   5x^2+11x+6
  •          (-) \ \   \underline{5x^2+5x}
  •                              6x + 6
  •                      (-) \ \   \underline{ 6x+6}
  •                                      \times

x^3+6x^2+11x+6 = (x+1)(x^2+5x+6) 

 = (x+1)(x^2+2x+3x+6) 

 = (x+1)[x(x+2)+3(x+2)] 

 = (x+1)(x+2)(x+3) 

Hence x^3+6x^2+11x+6 =(x+1)(x+2)(x+3)

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Question 8: Find the value of m , if mx^3+2x^2-3 and x^2-mx+4 leave the same remainder when each is divided by (x-2) .

Answer:

When x = 2

Remainder 1 = m(2)^3+2(2)^2-3 = 8m+5

Remainder 2 = (2)^2-m(2)+4 = 8-m

Given Remainder 1 = Remainder 2

8m+5=8-m \Rightarrow m = \frac{1}{3}

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Question 9: The polynomial px^3+4x^2-3x+q is completely divisible by (x^2-1) . Find the value of p \ and \ q . Also for these values of p \ and \ q , factorize the given polynomial completely.

Answer:

(x^2-1) is a factor of px^3+4x^2-3x+q

\Rightarrow (x-1)(x+1) is a factor of px^3+4x^2-3x+q

\Rightarrow (x-1) \ and \ (x+1) are factors of  px^3+4x^2-3x+q

When x = 1

p(1)^3+4(1)^2-3(1)+q = 0

\Rightarrow p+q=-1 … … … … … i)

When x = -1

p(-1)^3+4(-1)^2-3(-1)+q = 0

\Rightarrow p-q=7 … … … … … ii)

Solving i) and ii) we get p = 3 \ and \ q = -4

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Question 10: Find the number which should be added to x^2+x+3 so that the resulting polynomials completely divisible by (x+3) .

Answer:

Let a be added to the polynomial.

When x = -3

Remainder: (-3)^2+(-3)+3+a = 0 \Rightarrow a = -9

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Question 11: When the polynomial x^3+2x^2-5ax-7 is divided by (x-1) , the remainder is A . When the polynomial x^3+ax^2-12x+16 is divided by (x+2) , the remainder is B . Find the value of a  is 2A+B=0 .

Answer:

When x = 1

Remainder: (1)^3+2(1)^2-5a(1)-7=A

\Rightarrow 1+2-5a-7=A

\Rightarrow -5a-4=A … … … … … i)

When x = -2

Remainder: (-2)^3+a(-2)^2-12(-2)+16=B

\Rightarrow -8+4a+24+16=B

\Rightarrow 4a+32=B … … … … … ii)

Given 2A+B = 0

\Rightarrow 2(-5a-4)+(4a+32)=0

\Rightarrow -10a-8+4a+32=0

\Rightarrow a = 4

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Question 12: (3x+5) is the factor of the polynomial (a-1)x^3+(a+1)x^2-(2a+1)x-15 . Find the value of a and factorize the give polynomial.

Answer:

When x =-\frac{5}{3} 

Remainder: (a-1)(-\frac{5}{3})^3+(a+1)(-\frac{5}{3})^2-(2a+1)(-\frac{5}{3})-15=0 

-\frac{125}{27}(a-1)+\frac{25}{9}(a+1)+\frac{5}{3}(2a+1)-15=0 

-125(a-1)+75(a+1)+45(2a+1)-405=0 

a(-125+75+90)+125+75+45-405=0 

40a = 160 

\Rightarrow a = 4 

Hence (3x+5)  is a factor of  3x^3+5x^2-9x-15

  • 3x+5 ) \overline {3x^3+5x^2-9x-15} (x^2-3
  •  (-) \ \  \underline {3x^3+5x^2}  
  •                   -9x-15
  •          (-) \ \   \underline{-9x-15}
  •                           \times

3x^3+5x^2-9x-15 = (3x+5)(x^2-3) 

 = (3x+5)(x+\sqrt{3})(x-\sqrt{3}) 

Hence 3x^3+5x^2-9x-15 =(3x+5)(x+\sqrt{3})(x-\sqrt{3})

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Question 13: 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 14: 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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