G

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i was asked to factor 18+7x^2-x^4 and i factored it to (x^2+2)(x^2-9)

what did i do wrong????

G

i was asked to factor 18+7x^2-x^4 and i factored it to (x^2+2)(x^2-9)

what did i do wrong????

G

do what you can......ummmm........keep trying factor it out and ummmm...whatever

- Joined
- Feb 8, 2006

- Messages
- 22

You need one more step which is to simplify the factors:

(x^2-9) (x^2+2)....now you can finish it.

(x^2-9) (x^2+2)....now you can finish it.

Two things . . .I was asked to factor \(\displaystyle 18\,+\,7x^2\,-\,x^4\) and I factored it to \(\displaystyle (x^2\,+\,2)(x^2\,-\,9)\)

What did i do wrong?

(1) If you multiply out your answer, you get: \(\displaystyle \,x^4\,-\,7x^2\,-\,18\)

\(\displaystyle \;\;\;\)which is

(2) \(\displaystyle x^2\,-\,9\) can be factored, so you aren't finished yet.

I suspect you factored by taking out a -1 first: \(\displaystyle \;-1(x^4\,-\,7x^2\,-\,18)\)

Then we get: \(\displaystyle \:-1(x^2\,+\,2)(x^2\,-\,9)\)

Replace the\(\displaystyle \,-1:\;\;(x^2\,+\,2)(\underbrace{9\,-\,x^2})\)

. . . . . . . . . . . . . . . . . . . . diff. of squares

. . . . . Answer: \(\displaystyle \x^2\,+\,2)\overbrace{(3\,-\,x)(3\,+\,x)}\)

- Joined
- Feb 8, 2006

- Messages
- 6

in your first set of parqantheses it is a differrnce of two squares and you cannot further find the square of 2.

( x^2-9) is a difference of two squares .

since 2 does not have a square find the square of negative nine which does have a square

(x^2+2) (x-3) (x+3) you will end up with this outcome