Factorization of Cubic Polynomial


 
 
Concept Explanation
 

Factorization of Cubic Polynomial

Factorization of Cubic Polynomial: To factorise a cubic polynomial we perform the following steps:

  • Find a zero of the polynomial by hit and trial method
  • The perform long division of the polynomial by the factor obtained above
  • The quotient will be a polynomial of Degree 2
  • This quotient is then factorized by using the middle term spliting method.
  • Example: Factorise the following polynomial

    large x^3-6x^2+11x-6

    Solution: As it is a cubic polynomial one zero will be found using Factor Theorem

    Puting the value of x= 1 in the polynomial we get

    large f(1)=(1)^3-6(1)^2+11(1)-6

                large =1-6+11-6=0

    Therefore according to the Factor Theorem

    x-1 is factor of the given polynomial

    We will now perform long division with this factor

    large f(x)= x^3-6x^2+11x-6

                large = (x-1)(x^2-5x-6)

               large = (x-1)(x^2-6x+x-6)

               large = (x-1)[x(x-6)+1(x-6)]

               large = (x-1)(x-6)(x +1)

    Factorization of Bi-quadratic Polynomial: To factorise a bi-quadratic polynomial we perform the following steps:

  • Find two zeros of the polynomial by hit and trial method
  • Find the product of the two factors
  • The perform long division of the polynomial by the product of two factors obtained above
  • The quotient will be a polynomial of Degree 2
  • This quotient is then factorized by using the middle term spliting method.
  • Sample Questions
    (More Questions for each concept available in Login)
    Question : 1

    The number of zeroes in the polynomial      large x^3-2x^2+x-2    is ______________-

    Right Option : D
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    Question : 2

    On dividing large x^4+x^3+2x^2+3    by a polynomial   g(x),   we get the quotient    large x^2+x+7    and the remainder   5x  +  38.  The polynomial g(x) is

    Right Option : C
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    Explanation
     
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