How do you factor # x^4-8x^2-9#?

Answer 1

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

first lets substitute #x^2 = t# we get #t^2 - 8*t - 9# now split the middle term #t^2 + t - 9*t - 9# factorize now #t(t+1) -9(t+1)# now we get #(t+1)(t-9)# now substitute back #t=x^2# we get #(x^2+1)(x^2-9)# we can still factorize the #x^2-9# term so we get #(x^2+1)(x+3)(x-3)# this is the answer

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Answer 2

Isaiah Anthony

To factor the expression (x^4 - 8x^2 - 9), you can treat it as a quadratic in terms of (x^2). Let (u = x^2), then the expression becomes (u^2 - 8u - 9). Now, factor this quadratic expression. The factored form is ((u - 9)(u + 1)). Substituting back (x^2) for (u), you get ((x^2 - 9)(x^2 + 1)). Further factorization can be done using the difference of squares formula, resulting in ((x - 3)(x + 3)(x^2 + 1)). Therefore, the factored form of (x^4 - 8x^2 - 9) is ((x - 3)(x + 3)(x^2 + 1)).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.