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

Question

1626 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-05-15T19:19:28

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

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

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

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

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

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

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

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

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

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

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

How do you factor 8x3−4y2−50y+258x^3 - 4y^2-50y+25?