**x=2 is a root**of X*X*X =8 ...

**next divide**x^3-8

**by (x-2)**

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

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

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

x^{3} - 8

Note that if it had been x^{3} + 8, the factors would be (x+2)(x^{2}-2x+4)

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