How do you factor 8x^3 - 4y^2-50y+25?

May 15, 2015

I think the question is probably incorrect in mixing $x$ and $y$.

It looks like the intended problem was how to factor $8 {y}^{3} - 4 {y}^{2} - 50 y + 25$, which you can do by grouping:

$8 {y}^{3} - 4 {y}^{2} - 50 y + 25$

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

$= 4 {y}^{2} \left(2 y - 1\right) - 25 \left(2 y - 1\right)$

$= \left(4 {y}^{2} - 25\right) \left(2 y - 1\right)$

The $4 {y}^{2} - 25$ factor can be factored further.

Notice that $4 {y}^{2} - 25 = {\left(2 y\right)}^{2} - {5}^{2}$, which is of the form ${a}^{2} - {b}^{2}$.

${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$

So $4 {y}^{2} - 25 = {\left(2 y\right)}^{2} - {5}^{2} = \left(2 y + 5\right) \left(2 y - 5\right)$

So $8 {y}^{3} - 4 {y}^{2} - 50 y + 25 = \left(2 y + 5\right) \left(2 y - 5\right) \left(2 y - 1\right)$