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

Question

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

Answer 1

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 :)

Answer 2

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

(a³ - b³) = (a² +ab + b²)

In this case a = x since (x)³ = x³ and b = 2, since 2³ = 8.

 

So plug in x for a, and 2 for b into the following:

(a³ - b³) = (a² +ab + b²)

 

 

Giving you:

(x³ - 8) = (x-2)(x² + (x)(2) + 2²) = (x-2)(x² + 2x + 4)

 

Check

 

x(x² +2x + 4) - 2(x² + 2x + 4)

 

x³ +2x² +4x -2x² -4x - 8

 

The x² terms and the x terms cancel out and you are left with:

 

x³ - 8

 

Note that if it had been x³ + 8, the factors would be (x+2)(x²-2x+4)