Question

I really could use some help with Factoring a trinomial problem... Anybody able to do that?

Answer 1

What is it?

Answer 2

2x^4+11x^2+12

Answer 3

Substitute x^2 = y. You get:
2y^2 + 11y + 12. Factor that, and then substitute back for y.

Answer 4

I dont understand why I am substituting a variable for another variable

Answer 5

Quadratic factoring is easier. :)

Answer 6

Okay... I don't know how to do that. I understand substituting but I don't know how to do that.

Answer 7

(2y+3)(y+4)
(2x^2 + 3)(x^2 + 4)

Answer 8

okay but when you add the 3 and the 4, it does not equal 11

Answer 9

So I guess I am just wondering how you came up with the 3 and the 4 that you are adding to x

Answer 10

Well, 3*4 is 12. That works for the constant.
3y + 8y = 11y. That works for that term.
2y^2 works for the first term.

It is just a matter of finding right numbers (factors) and rearranging.

Answer 11

Okay so I am gonna try my next problem... do you mind hanging on and maybe helping me with that one too?

Answer 12

I can.

Answer 13

Thanks... you are just such a life saver!

Answer 14

Okay so I have y^4-11y^3+28y^2

Answer 15

y^2(y-11)(y-4)?

Answer 16

I lied... y^2(y-7)(y-4)

Answer 17

y^2 is a factor. So, you get y^2 (y^2 - 11y + 28).

Now, to factor the rest, you need to find two numbers that multiply to 28 and add up to 11. Can you think of any?

Answer 18

Good. Your answer is right.

Answer 19

YAY!

Answer 20

Are you still around

Answer 21

now i am