Question

Hallie subtracted a quantity from the polynomial 3y^2 + 8y - 16 and produced the expression (y + 2)(y - 2). What quantity did Hallie subtract? Explain how you got your answer.

Answer 1

I will try to do the work with you...
We have the starting polynomial (the one that we will subtract from) to be \(3y^2 + 8y - 16\) and the unknown to be \(x\). The difference will be \((y+2)(y - 2)\).
\(3y^2 + 8y - 16 - x = (y + 2)(y - 2)\).

I would suggest for you to expand from the factorized form of \((y+ 2)(y - 2)\) by using \(\text {FOIL}\).

Answer 2

Make \((y+2)(y-2)\) a polynomial by using FOIL. Do you know how to do that?

Answer 3

Yes. After doing that I would set it equal to the one you mentioned, and then I would simplify, right?

Answer 4

If you mean to isolate \(x\) and find the value of it, then yes.

Answer 5

Oh yeah, I got it. Thanks! I'll let you know what I got just to make sure.

Answer 6

Okay

Answer 7

@AmberlyKhan Did you get it?

Answer 8

Is it x = 12 + 2y^2 - 8y?

Answer 9

And I would factor that?

Answer 10

I did something wrong, because the difference I got was y^2 - 28. It shoud be y^2 - 4.

Answer 11

Let's check:
\((y+2)(y-2) = y^2 - 2y + 2y - 4 = y^2 - 4\)

\((3y^2 + 8y - 16) - x = y^2 - 4\)
\((3y^2 + 8y - 16) - (3y^2 + 8y - 16) - x = (y^2 - 4) - (3y^2 + 8y - 16)\)
\(-x = y^2 - 4 - 3y^2 - 8y + 16\)
\(-x = - 2y^2 - 8y + 12\)
\(x = 2y^2 + 8y - 12\)

Answer 12

How'd you get +16? I got -16.

Answer 13

There's an invisible -1 in front of \((3y^2 + 8y - 16)\). So it's basically like you're distributing -1 to -16 to get 16.

Answer 14

Oh, nevermind, I got it.

Answer 15

You distributed the negative.

Answer 16

Yes I distributed the -1

Answer 17

Do you understand this now?

Answer 18

OK, I got the answer. Thanks for bearing with me!