# How do you factor 5y^2-7y-6?

Mar 27, 2018

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

#### Explanation:

I made the assumption that all coefficients and factors were integers. Based on this, I started building a form of the factored equation:

$\left(a y + c\right) \left(b y + d\right) = a b {y}^{2} + \left(a d + b c\right) y + \mathrm{dc}$

The expanded form allows us to make some guesses, again assuming integers.

We know that $a b = 5$ which means a or b has to equal 1 or 5. Let's plug that in:

$\left(5 y + c\right) \left(y + d\right) = 5 {y}^{2} + \left(5 d + c\right) y + \mathrm{dc}$

Next, we need to figure out what d and c are. We know that $\mathrm{dc} = - 6$ which gives us two integer options. Either d&c are 2&3, or they are 1&6.

The trick here is to figure out the values of d/c using the middle term:

$5 d + c = - 7$

If $d = - 2$ and $b = 3$, it satisfies the middle equation:

$5 \left(- 2\right) + 3 = - 7 \Rightarrow - 10 + 3 = - 7$

Now we know all of our variables, and can write out the final factored equation:

$\textcolor{red}{\left(5 y + 3\right) \left(y - 2\right)}$