# How do you factor completely: 4x^3 - 8x^2 - 26x + 50?

Jul 23, 2015

As written, it cannot be factored by the methods taught in pre-caculus level algebra classes. If the 26 was supposed to be 25, then factor by grouping.

#### Explanation:

If you can find the roots using the cubic formula (nost of us do not have that formula memorized), then you can factor.

If there is an eror in the question, and the intended question is:

How do you factor completely: $4 {x}^{3} - 8 {x}^{2} - 25 x + 50$?

Then you can factor by grouping (and the difference of squares):

$4 {x}^{3} - 8 {x}^{2} - 25 x + 50 = \left(4 {x}^{3} - 8 {x}^{2}\right) + \left(- 25 x + 50\right)$

$= 4 {x}^{2} \left(x - 2\right) + \left(- 25\right) \left(x - 2\right)$

$= \left(4 {x}^{2} - 25\right) \left(x - 2\right)$

$= \left(2 x + 5\right) \left(2 x - 5\right) \left(x - 2\right)$