how does this get changed in to (x-2)(x^2+2x+4) ?

## x^3-8 How to change this to (x-2)(x^2+2x+4) ?

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# 2 Answers

Hey Jo -- we may notice

**x=2 is a root**of X*X*X =8 ...**next divide**x^3-8**by (x-2)**=> x^2 +2x +4 ... factors are thus (x-2)(x^2 +2x +4) ... Best wishes :)

This is the difference of two cubes. It is similar to a difference of two squares in that there is a formula for factoring it.

This is the formula

(a

^{3}- b^{3}) = (a^{2}+ab + b^{2})In this case a = x since (x)

^{3}= x^{3 }and b = 2, since 2^{3}= 8.So plug in x for a, and 2 for b into the following:

(a

^{3}- b^{3}) = (a^{2}+ab + b^{2})Giving you:

(x

^{3}- 8) = (x-2)(x^{2}+ (x)(2) + 2^{2}) = (x-2)(x^{2}+ 2x + 4)Check

x(x

^{2}+2x + 4) - 2(x^{2}+ 2x + 4)x

^{3}**+2x**+4x^{2}**-2x**-4x - 8^{2}The x

^{2}terms and the x terms cancel out and you are left with:x

^{3}- 8Note that if it had been x

^{3}+ 8, the factors would be (x+2)(x^{2}-2x+4)