Question

factor completely: 4y^3+10y^2+12y+30
i factored out the 2 and got 2(2y^3+10y^2+6y+15)

Answer 1

let's group it first

let's grab 2 guys and then another 2 guys so

\(\bf
4y^3+10y^2+12y+30\\
(4y^3+10y^2)+(12y+30)\)

can you get common factor off each pair there?

Answer 2

so it would be \[2y^2(2y+5) + 3(4y+10)\] i think

Answer 3

well, the 1st pair seems to not have any more common factors
but the 2nd pair does

Answer 4

4y + 10, have a common factor still

Answer 5

2

Answer 6

right, so get that one too :)

Answer 7

so would i times the 3 and the 2 together?

Answer 8

yes, they'd multiply each other

Answer 9

it'd have been the same as if you had used "6" instead of 3 at first

Answer 10

so it would be \[2y^2(2y+5)+ 6(2y+5) \] which would then be \[(2y^2+6)(2y+5)\] ?

Answer 11

yes :)

Answer 12

thank you!!!!

Answer 13

yw