# Thread: can some one help me with factoring

1. ## can some one help me with factoring

factoring completely

250x3-128y3

x2+100

2. Originally Posted by dblued
factoring completely

250x3-128y3
$\displaystyle 250x^3 - 128y^3 = 2(125x^3 - 64y^3) = 2[(5x)^3 - (4y)^3]$

now use the difference of two cubes formula: $\displaystyle a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

here of course, your $\displaystyle a$ is $\displaystyle 5x$ and your $\displaystyle b$ is $\displaystyle 4y$

3. Originally Posted by dblued
factoring completely

250x3-128y3

x2+100
$\displaystyle 250x^3 - 128y^3 = 2(125 x^3 - 64 y^3) = 2([5x]^3 - [4x]^3)$. Now apply the formula for the difference of two perfect cubes.

$\displaystyle x^2 + 100 = (x + 10i)(x - 10i)$ where i is the square root of -1.

4. Originally Posted by Jhevon
$\displaystyle 250x^3 - 128y^3 = 2(125x^3 - 64y^3) = 2[(5x)^3 - (4y)^3]$

now use the difference of two cubes formula: $\displaystyle a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

here of course, your $\displaystyle a$ is $\displaystyle 5x$ and your $\displaystyle b$ is $\displaystyle 4y$
Snap

5. Originally Posted by dblued
x2+100
did you mean $\displaystyle x^2 ~{\color{red}-}~ 100$?

maybe that's too easy

6. Originally Posted by mr fantastic
$\displaystyle x^2 + 100 = (x + 10i)(x - 10i)$ where i is the square root of -1.
i thought about doing that, but complex numbers seemed, well, too advanced here...