Question

Factor 6x^4 - 4x^2 + 8x
Factor 8p^3 + 12p^2 + 4p
Factor 12k^3 + 6k^2 - 18k It would be helpful if you show how you get the answers, please and thank you very much.

Answer 1

the gcf of 6x^4,4x^2,8x is 2x

so we have
2x(3x^3-2x+4)

Answer 2

im aware of all the GCFs but how do i factor them afterwards

Answer 3

it is factored

Answer 4

we just factored it i mean

Answer 5

No... How do you factor the problem completely

Answer 6

8p^3 + 12p^2 + 4p
=4p(2p^2+3p+4) GCF
the trinomial in this one and the one that m'naya did do not factor over the integers

Answer 7

oh okay so that is the final answer

Answer 8

you can try syntheic division to find a zero of 3x^3-2x+4

Answer 9

but that only works if it has rational roots

Answer 10

12k^3 + 6k^2 - 18k
=6k(2k^2+k-3) GCF
=6k(2k+3)(k-1)
there is a method for factoring the trinomial, but since 2 and 3 are prime trial and error is what i used; i'll show you if you want

Answer 11

please do..

Answer 12

ac=2*(-3)=-6
then factors of ac that sum to b are
-2*3=-6 and -2+3=+1
so we can use -2 and 3 as coefficients to rewrite the +k term

=2k^2 -2k +3k -3

then we factor by grouping (take GCF from first two and last two terms)

=2k(k-1)+3(k-1)

to finish factoring by grouping we factor out the k-1 as a GCF

=(k-1)(2k+3)

Answer 13

a, b, and c refer to \[ax^2+bx+c\]

Answer 14

You know that factoring and multiplication are inverse operations. the method takes you to the step following a FOIL:

(k-1)(2k+3)=2k^2+3k-2k-3

compare the above line to my previous post; they are the same except that the middle two terms are commuted.