1. ## Factor each polynomial

I need help with these three questions,
can someone explain to me CLEARLY how to factor each polynomial

1. 125x^6 - 8
2. 64x^12 - 27
3. 8/125x^3 - 64y^6

2. Originally Posted by skeske1234
I need help with these three questions,
can someone explain to me CLEARLY how to factor each polynomial

1. 125x^6 - 8
2. 64x^12 - 27
3. 8/125x^3 - 64y^6
Hi skeske1234,

Do you know the model for factoring the difference of two cubes?

Review this:

$\displaystyle x^3-y^3=(x-y)(x^2+xy+y^2)$

Apply that to your problems. First let's make them look a bit more like the difference of two cubes.

[1] $\displaystyle (5x^2)^3-(2)^3$

[2] $\displaystyle (4x^4)^3-(3)^3$

[3] $\displaystyle \left(\frac{2}{5}x\right)^3-(4y^2)^3$

3. Originally Posted by masters
Hi skeske1234,

Do you know the model for factoring the difference of two cubes?

Review this:

$\displaystyle x^3-y^3=(x-y)(x^2+xy+y^2)$

Apply that to your problems. First let's make them look a bit more like the difference of two cubes.

[1] $\displaystyle (5x^2)^3-(2)^3$

[2] $\displaystyle (4x^4)^3-(3)^3$

[3] $\displaystyle \left(\frac{2}{5}x\right)^3-(4y^2)^3$
after you do that, how do you plug it into the formula?

4. Originally Posted by masters
Hi skeske1234,

Do you know the model for factoring the difference of two cubes?

Review this:

$\displaystyle x^3-y^3=(x-y)(x^2+xy+y^2)$

Apply that to your problems. First let's make them look a bit more like the difference of two cubes.

[1] $\displaystyle (5x^2)^3-(2)^3$

[2] $\displaystyle (4x^4)^3-(3)^3$

[3] $\displaystyle \left(\frac{2}{5}x\right)^3-(4y^2)^3$
Originally Posted by skeske1234
after you do that, how do you plug it into the formula?
I'll do the first one, and you see if that helps you do the other two.

$\displaystyle x^3-y^3=(x-y)(x^2+xy+y^2)$

In my model, let's let $\displaystyle x=5x^2 \ \ and \ \ y=2$

$\displaystyle (5x^2-2)\bigg((5x^2)^2+(2)(5x^2)+(2)^2\bigg)$

$\displaystyle (5x^2-2)(25x^4+10x^2+4$

### what is the of factor of 125x^6y^6-8

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