SOLUTION: Factor trinomials. 2x^2+10x+8 2(x^2+5x+4) 2(x^2+1)(5x+4) Is this right?

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Question 139245: Factor trinomials.
2x^2+10x+8
2(x^2+5x+4)
2(x^2+1)(5x+4)
Is this right?

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
Let's take your answer and multiply it back out.

2%28x%5E2%2B1%29%285x%2B4%29=2%285x%5E3%2B4x%5E2%2B5x%2B4%29=10x%5E3%2B8x%5E2%2B10x%2B8. Hmm, doesn't look like what you started with.

2%28x%5E2%2B5x%2B4%29

Both signs in the trinomial are +, so both of our factors must have a +, and since we factored out the 2, our lead coefficient on the trinomial is 1, so that means the first term of each of our binomial factors is x. So far we have:

(x + _)(x + _)

The numbers that go into the two blanks have to be 4 when you multiply them together and 5 when you add them. Hmmm, 4 times 1 is 4 and 4 plus 1 is 5. Looks good, so:

(x + 1)(x + 4)

Now put the 2 back:
2(x + 1)(x + 4)

And you are done.

Check:
x times x is x%5E2
1 times x is x
x times 4 is 4x
1 times 4 is 4
Add 'em up:
x%5E2%2B5x%2B4
And put the 2 back in:
2%28x%5E2%2B5x%2B4%29=2x%5E2%2B10x%2B8 Answer checks.