SOLUTION: When x^4-8x^2-9 is factored as completely as possible, which of the following is a prime factor ? a) (x+1) b) (x^2-9) c) (x+1)^2 d) (x+3) e) (x^2+9) I don't understand it

Question 822853: When x^4-8x^2-9 is factored as completely as possible, which of the following is a prime factor ?
a) (x+1)
b) (x^2-9)
c) (x+1)^2
d) (x+3)
e) (x^2+9)
I don't understand it and can someone explain how to go about doing this problem?
Thanks, much is appreciated.
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
The goal is to factor. So we must find two numbers that multiply to -9 (last term) AND add to -8 (middle coefficient). These two numbers are +1 and -9, so we break up the middle term -8x^2 into +1x^2 - 9x^2 and we factor by grouping like so...

x^4-8x^2-9

x^4+x^2-9x^2-9

(x^4+x^2)+(-9x^2-9)

x^2(x^2+1)+(-9x^2-9)

x^2(x^2+1)-9(x^2+1)

(x^2-9)(x^2+1)

(x^2-3^2)(x^2+1)

(x-3)(x+3)(x^2+1) ... use the difference of squares rule here

---------------------------------------------------------------------------

x^4-8x^2-9

completely factors to

(x-3)(x+3)(x^2+1)

So the final answer (that they want) is d) (x+3)

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