Question

How do you factor completely: `10x^5 + 4x^4 + 8x^3`?

Answer 1

`10x^5 + 4x^4 + 8x ^3` `= 2 x^3 ( 5 x^2 + 2 x + 4)` `= 10 x^3 (x - 1/ 5 ( -1 + i\sqrt{19}))(x - 1/ 5 ( -1 - i\sqrt{19}))`

Explanation:

`10x^5 + 4x^4 + 8x ^3`

First we take out the obvious common factor of `x^3`. Let's take out the slightly less obvious factor of `2` as well.

`= 2 x^3 ( 5 x^2 + 2 x + 4)`

That's pretty good. If we can factor the quadratic equation we can make more progress. To check if we can, we can check if the discriminant `b^2-4ac` is a perfect square. Well `2^2-4(5)(4)=-76` is a negative number, which won't ever be a perfect square.

Depending on what grade we're in, we either stop here or factor using complex numbers. I'll continue.

Pro tip: The Shakespeare Quadratic Formula (`2b` or `-2b`) says `x^2-2bx+c` has zeros `x= b\pm sqrt{b^2-c}` and `ax^2+2bx+c` has zeros `x = \frac 1 a(-b \pm sqrt{b^2 - ac})`.

By the Shakespeare Quadratic Formula the quadratic has zeros

`x = 1/ 5 ( -1 \pm \sqrt{-19}) = 1/5 (-1 pm i \sqrt{19})`

so we can factor

`5 x^2 + 2 x + 4`

`= 5(x - 1/ 5 ( -1 + i\sqrt{19}))(x - 1/ 5 ( -1 - i\sqrt{19}))`

and the entire expression

`10x^5 + 4x^4 + 8x ^3`

`= 10 x^3 (x - 1/ 5 ( -1 + i\sqrt{19}))(x - 1/ 5 ( -1 - i\sqrt{19}))`