Question

Factor the trinomial: x^2-10x+25-y^2

Answer 1

That is supposed to be \(x^2\) right?

Answer 2

Ohh, yes. Sorry about that. I'll fix it.

Answer 3

Ok, I see.

So the first step is to factor the x's. That actually factors pretty nicely. Can you do it, or do you want me to show you?

Answer 4

Well, if you can show me how, that would be awesome. I'm not clear on the concept of this. I never thought about just factoring out x's because not every term has an x.

Answer 5

\[x^2 - 10x + 25 - y^2\]\[(x - 5)^2 - y^2\]

Answer 6

If you assume that \(z = x - 5\) then we can factor \(z^2 - y^2\) which is also easy to factor.

Answer 7

Ahh, so you use substitution?

Answer 8

That's how I'd do it anyway.

Then you end up with \((z + y)(z - y)\) and you can just substitute back.

Answer 9

Okay, that's making more sense than whatever caterwompus info I found in my online text. Thank you!

Answer 10

BTW did @hallerdonald ever find his question that you were answering?

Answer 11

No idea. He never replied.

Answer 12

Lolol. Well he told me he was new. He, for some reason, thought my question was the answer to his... ?? I told him to message you because you were replying but, IDK what TH he did.

Answer 13

Oh, well, I don't know then. :D

Anyway, I'll also throw out there that you can "pull apart" constants to make this work even if nothing immediately jumps out at you. Like if it was +29 and there was a -2y or something, you could keep the 25 where it is and the +4 in with the y. You know?

Answer 14

Sort of... Let me read that again.

Answer 15

I'll write it out, so it's hopefully more clear.

Let's say we had to factor \[x^2 + y^2 - 10x - 2y + 29\]
That looks ugly at first. But we could rewrite that as
\[x^2 - 10x + 25 + y^2 - 2x + 4\]And by doing that get two nicely factorable parts.

Answer 16

Oops, -2y in the second thing. But yeah.

Answer 17

Ahh, I see what you're saying now.

Answer 18

Thank you for pointing that out. >>>>