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

Question

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

jdoe0001 · Answer

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?

bekkah323 · Answer

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

jdoe0001 · Answer

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

jdoe0001 · Answer

4y + 10, have a common factor still

bekkah323 · Answer

2

jdoe0001 · Answer

right, so get that one too :)

bekkah323 · Answer

so would i times the 3 and the 2 together?

jdoe0001 · Answer

yes, they'd multiply each other

jdoe0001 · Answer

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

bekkah323 · Answer

so it would be $$2y^2(2y+5)+ 6(2y+5) $$ which would then be $$(2y^2+6)(2y+5)$$ ?

jdoe0001 · Answer

yes :)

bekkah323 · Answer

thank you!!!!

jdoe0001 · Answer

yw