Question

factor: 2x^3 + 6x^2 + 2x + 6

Answer 1

what can be taken out of ALL of these terms? are all of these numbers divisable by 2?

Answer 2

Yes then I get confused

Answer 3

?

Answer 4

it's okay :) so

we have 2(x^3 + 3x^2 + x + 3) right

dont worry about the +3 for now

an x can be taken out of x^3, 3x^2 and x right?
so

2x(x^2 + 3x + 1) is what we're left with right?

can (x^2 + 3x + 1) be factored out? no so it'll have to stay...so just add back in the +3

so we have

2x(x^2 + 3x + 1) + 3

multiply it out to check if you're unsure

Answer 5

Ok thanks :) this helped so much

Answer 6

no problem :) and ill double check to make sure that's correct before i leave this question

Answer 7

None of the answer choice look like this thought they are
(2x + 6)(x^2 + 1)

2[(x + 3)(x^2 + 1)]

(x + 3)(2x^2 + 2)

Answer 8

and prime

Answer 9

hang on i see my mistake

Answer 10

ok thanks for doing all of this :)

Answer 11

hang on computer is acting up

Answer 12

ok :)

Answer 13

there we go...sorry it was updating haha..too slow!

okay

here's where i went wrong..ill start over and point it out

we factored out the 2 right

and had

2(x^3 + 3x^2 + x + 3) right?

Answer 14

yes

Answer 15

okay...what you do next...with these quadnomials...is you group them into binomials....

you take

(x^3 + 3x^2) + (x + 3)

see how we grouped them? from 4 terms in 1 parenthesis....to 2 terms in 2 parenthsis....okay?

Answer 16

yes

Answer 17

I think I just got it!

Answer 18

okay...so....now we can factor again...what can come out of the left parenthesis...and what (if anything...can come out of the right?)

Answer 19

what'd you get?

Answer 20

2( x + 3)(x^2 + 1)

Answer 21

Thanks for helping

Answer 22

:) very good :) and you're very welcome