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
Algebra ->
Polynomials-and-rational-expressions
-> 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
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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
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