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Thread: 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

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

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

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


    Therefore

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

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

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

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

    Comparing constants, $\displaystyle c=4$

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

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

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