# Math Help - Factoring Trinomials

1. ## Factoring Trinomials

My problem states to factor completely:

49c+56c+16

Nothing can be factored out as it is. So I took the product of 49 & 16 = 784 and two integers whose products equal 784 and sum equals 56 is (28)(28)

Yet when I go to factor, it's not adding up. Maybe I am confusing the steps?

49c2+28c+28c+16

28c(c+21c) ?

I am not sure what I am suppose to do actually...

4. ## Re: Factoring Trinomials

Hello, Kibbygirl!

Factor completely: . $49c^2+56c+16$

Nothing can be factored out as it is.
So I took the product of 49 & 16 = 784
. . and two integers whose products equal 784 and sum equals 56 is (28)(28).

Yet when I go to factor, it's not adding up. Maybe I am confusing the steps?

$49c^2+28c+28c+16$ . This is correct!

28c(c+21c) ? .But what is this?

You are supposed to "factor by grouping".
Are you familiar with it?

You have: . $49c^2 + 28c + 28c + 16$

Factor the terms "by pairs".
. . You can factor $7c$ from the first two terms
. . You can factor $4$ from the last two terms.

Then you have: . $7c(7c+4) + 4(7c+4)$

Note that the term groups have $(7c+4)$ on common . . . factor it out!

And we have: . $(7c+4)(7c+4) \:=\:(7c-4)^2$

See how it works?

5. ## Re: Factoring Trinomials

Thank you! I didn't realize I should be factor by grouping, I was trying to do the form x2+bx+c. I am vaguely familiar with factoring by grouping, so thank you for the steps.

6. ## Re: Factoring Trinomials

Soroban, can you tell me why in your final it is - and not +? Everything I am seeing in my book shows it'd be (7c+4)2??

7. ## Re: Factoring Trinomials

Originally Posted by Kibbygirl
Soroban, can you tell me why in your final it is - and not +? Everything I am seeing in my book shows it'd be (7c+4)2??
That was just a typo.

8. ## Re: Factoring Trinomials

Ok great, just wanted to verify in case I am not understanding correctly.