Question

Factor the polynomial;
16j(squared)k-8j(to the sixth power)k(to the fifth power)+60j(to the third power) *sorry guys don't know how to type numbers with powers! :/

Answer 1

to type something like \(x^2\), you type x^2

you use the ^ symbol, found above the 6 (you have to hit shift, then 6)

Answer 2

so

16j(squared)k

shortens to

16j^2k

Answer 3

So all together, the problem is

16j^2k - 8j^6k^5 + 60j^3

Answer 4

oh okay gotcha!! thanks a lot!(: Now I need help solving that! lol

Answer 5

the GCF of the coefficients is 8

the GCF of the variable terms is j^2

so overall, the GCF is 8j^2

Answer 6

factor out 8j^2

Answer 7

and tell me what you get

Answer 8

oh i'm sry, it's not 8j^2, I just saw the 60 and I'm not thinking

it should be 4j^2

Answer 9

Yeah, its okay! I was so confused for a second there! But i thought my answer would be 4j^2...

Answer 10

yeah sorry about that, you would factor out the GCF 4j^2 to get

16j^2k - 8j^6k^5 + 60j^3

4j^2*4k - 4j^2*2j^4k^5 + 4j^2*15j

4j^2(4k - 2j^4k^5 + 15j)

Answer 11

That's just a little confusing lol!

Answer 12

basically what I did was factor each individual expression into two factors

one of the factors is 4j^2

Answer 13

then I used the distributive property

Answer 14

notice if you distribute the 4j^2 back

you'll go from

4j^2(4k - 2j^4k^5 + 15j)

to

16j^2k - 8j^6k^5 + 60j^3

Answer 15

oh okay, now it makes sense!

Answer 16

ok that's great, glad it does

Answer 17

Yeah, Thanks a lot!! (:

Answer 18

yw