How do you factor 2y^2 +5y-18?

1 Answer
Mar 4, 2018

(2y + 9)(y-2)

Explanation:

We can use the "AC Method" (at least that was the name of the method, which I will explain further, that I was taught) to factor the quadratic equation 2y^2 + 5y - 18.

Why is it called the "AC Method," you may ask?

In the standard quadratic equation ax^2 + bx + c = y, we are going to be first working with the a and the c coefficients. (Just replace the x's with y's -- variables can be changed for the purposes of understanding the concept pertaining to your question.)

The a and the c coefficients in the equation you posted are 2 and -18, respectively.

So in this method, what we do first is multiply the a and the c together: 2(-18) = -36.

Then we find factors that will result in the product of -36 but when they are added together will result in the sum of the b coefficient (which is 5 in this case).

Aha! We have found -4 and 9 because -4(9) = -36 and -4 + 9 = 5.

Now we expand and write an unsimplified version of the equation.
2y^2 - 4y + 9y - 18 =0

Now we factor by grouping...
2y(y - 2) and 9(y-2)

Notice I just paired two terms in the expanded form of the equation and factored through GCF (Greatest Common Factor). Also, notice the sharing of the expression y-2 (this is extremely important).

Put 2y and 9 together into (2y + 9) and we already have (y-2).

That's it! Our factored equation is (2y + 9)(y-2). Multiply it out and you should hopefully get back to 2y^2 + 5y - 18.

*I suggest you look for videos for this method because explaining it through written words, speaking from my experience, did not help me fully grasp the concept. *