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Math Help - How to factor (x^3 + 8) ?

  1. #1
    s3a
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    How to factor (x^3 + 8) ?

    How do I factor (x^3 + 8) ?

    Any help would be greatly appreciated!
    Thanks in advance!
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  2. #2
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    Quote Originally Posted by s3a View Post
    How do I factor (x^3 + 8) ?

    Any help would be greatly appreciated!
    Thanks in advance!
    Notice that

    x^3 + 8 = x^3 + 2^3, a sum of two cubes.

    So it can be factorised using the sum of two cubes rule...

    a^3 + b^3 = (a + b)(a^2 - ab + b^2).


    Therefore

    x^3 + 8 = (x + 2)(x^2 - 2x + 4).
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    Let f(x)=x^{3}+8

    Since f(-2)=0, by factor theorem, (x+2) is a factor.

    So x^{3}+8=(x+2)(ax^{2}+bx+c)

    Comparing coeff. of x^{3}, a=1

    Comparing constants, c=4

    Comparing coeff. of x^{2}, b+2a=0
    b=-2

    Note: you can find the quadratic factor by long division also.

    Thus
    f(x)=(x+2)(x^{2}-2x+4)
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