Question

Factor the following completely Im not sure how to do this please show work I will give medals
1. c^2 – 121
2. p^2 + 22p + 40

Answer 1

I told you how to do both of these when you posted them before. Please do not repost if you aren't going to heed what you are told.

Answer 2

sorry i was still confused

Answer 3

Then ask questions!

Here's the first one:

\[a^2-b^2 = (a-b)(a+b)\]

Compare \[a^2-b^2\]\[c^2-121\]what is the square root of 121? Are those expressions not equivalent if \(a=c\) and \(b = \sqrt{121}\)?

Answer 4

For the second one, use the splitting or grouping technique I showed you earlier:

\[p^2 + 22p + 40\]Multiply the coefficient of the first and last terms: 1*40 = 40
Now choose a pair of factors of 40 that add to 22: 2*20 = 40, 2+20 = 22

Rewrite the polynomial, splitting the middle term:\[p^2 + 20p + 2p + 40\]Now group each pair\[(p^2+20p) + (2p+40)\]Now factor each group; what do you get?

Answer 5

so the first ones answer would be (c - 11)(c + 11)?

Answer 6

Yes!

Answer 7

Whenever you see <something> - <something else>, you should always be investigating whether both things are squares like this. It's just about the easiest factoring of all.

Answer 8

and the second ones answer would be (p + 20)(p + 2)?

Answer 9

Yes again!

\[(p^2+20p) + (2p+40)\]\[p(p+20) + 2(p+20)\]\[(p+20)(p+2)\]

Checking our work:
\[(p+20)(p+2) = p^2 + 2p + 20p + 20*2 = p^2 + 22p + 40\checkmark\]

Answer 10

Checking the other one, for good measure:
\[(c+11)(c-11) = c^2 -11c + 11c -121 = c^2 -121\checkmark\]

Answer 11

thank you

Answer 12

You're welcome. I hope you're now a factoring master :-)