Question

Completely factor the polynomial below.
16q^2+20q+6
A. (8q+3)(2q+1)
B. (8q+1)(2q+3)
C. 2(4q+3)(2q+1)
D. 2(4q+1)(2q+3)

Answer 1

Did you attempt to factor it?

Answer 2

I am working on that as we speak

Answer 3

16q^2 + 20q +6
= 16^q^2 + 12q+8q +6
= 4q(4q+3) +2(8q+3)
= (4q+3)(4q+2)ans

Answer 4

You can pull a two out before you starting doing FOIL

Answer 5

Because that is the factor they have in common right?

Answer 6

Yes, 20, 16, and 6, all have a common factor of 2.

Answer 7

How would you factor the 2s out first

Answer 8

\[16q^2+20q+6 = 2(8q^2 + 10q + 3)\]

Answer 9

Does that make sense to you?

Answer 10

To be completely honest no because I don't know what you would do next.

Answer 11

When you factor out first

Answer 12

You start with the factors of 8, which are 8, 4, 2, and 1.

\[2(q+ 1)(q + 3) \]

You try out different coefficients (numbers in front of variables) for q. There are only two factors for 3, so you know it is 1 and 3. The positioning of these numbers may change depending on the coefficients you use.

Answer 13

\[(Kq + 1)(Zq + 3) = 8q^2 + 10q + 3\]

Just try different coefficient values for K and Z, then foil them out to see which one gets that polynomial. A tip is to start with the smaller numbers (for the factors of 8) which is 2 and 4.

Let's try:
\[(4q + 3)(2q + 1) \]
Try FOIL it and see if that works

Answer 14

When I said smaller numbers, what I was going for was numbers that are closer together. Usually when the terms are closer in value, they are the correct answer.