# How do you factor 10y^2-31y+15  ?

Use a table to best see the different pieces of the factorization and you'll get to $\left(2 y - 5\right) \left(5 y - 3\right)$

#### Explanation:

Working these requires a bit of patience.

Finding the two expressions that will factor this completely involves finding numbers that will add to -31 but multiply to 10 and 15.

We do know that the 31 is negative, so there are negative values in this factoring exercise. However, 15 is positive, so there will be an even number of negatives that when multiplied together will get us to a positive number.

So what we know so far is that the factors will look like:

$\left(a y - b\right) \left(c y - d\right)$

where $a c = 10 , b d = 15 , \mathmr{and} a d + b c = - 31$

Let's table this out and see what we find. I'm just going to try values until I find ones that work:

$a = 2 , c = 5 , b = - 3 , d = - 5 , a d + b c = - 10 - 15 = - 25$
$a = 2 , c = 5 , b = - 5 , d = - 3 , a d + b c = - 6 - 25 = - 31$

And those work! Let's put those into our format:

$\left(2 y - 5\right) \left(5 y - 3\right)$