help on factoring`

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i got this problem wrong and i did not know what i did wrong....

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????
 
Hello!

do what you can......ummmm........keep trying factor it out and ummmm...whatever
 
18+7x^2-x^4

Multiply by -1

x^4-7x^2-18

=(x^2-9)(x^2+2)


Factor, Difference of two squares x^2+-9

(x-3)(x+3)

Answer


(x^2+2)(x-3)(x+3)
 
Hello, brendaursula!

I was asked to factor 18+7x2x4\displaystyle 18\,+\,7x^2\,-\,x^4 and I factored it to (x2+2)(x29)\displaystyle (x^2\,+\,2)(x^2\,-\,9)

What did i do wrong?
Two things . . .

(1) If you multiply out your answer, you get: x47x218\displaystyle \,x^4\,-\,7x^2\,-\,18
      \displaystyle \;\;\;which is not what they gave you.

(2) x29\displaystyle x^2\,-\,9 can be factored, so you aren't finished yet.


I suspect you factored by taking out a -1 first:   1(x47x218)\displaystyle \;-1(x^4\,-\,7x^2\,-\,18)

Then we get: 1(x2+2)(x29)\displaystyle \:-1(x^2\,+\,2)(x^2\,-\,9)

Replace the1:    (x2+2)(9x2)\displaystyle \,-1:\;\;(x^2\,+\,2)(\underbrace{9\,-\,x^2})
. . . . . . . . . . . . . . . . . . . . diff. of squares
. . . . . Answer: \(\displaystyle \:(x^2\,+\,2)\overbrace{(3\,-\,x)(3\,+\,x)}\)
 
ok, you have(x^2+2) (X^2-9).
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
 
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