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Determine the zeros of f(x) = x4 - x3 + 7x2 - 9x - 18.

Mathematics
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-1 fits in.
Got it, break 7 as 9-2 /
so \[\pm 1\] is the answer?

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Other answers:

nopes.
What do get after writing 7 as 9-2 ?
\[x^4 - x^3 + 9 - 2x^2 - 9x -18\] ?
x^4 - x^3 + 9x^2 -2x^2 - 9x - 18.=0 =>(x^4 + 9x^2) - (x^3 -9x) -(2x^2 -18) =0 You see what should be the next step ?
yeah, factor each binomial, right?
yep.
(x^2 + 1) (x^2 + 9) for the first one?
try again.
I'm confused
x1 = 2 x2 = -1 x3 = 0 + 3i x4 = 0 - 3i
So, substitute (0 - 3i) and -1 for x^2, making it ((0 - 3i) + 9(-1))?
so, then it would become (-3i - 9)
hint : divide f(x) to x^2+9 you would find your answer!
So, divide x^4 - x^3 + 7x^2 - 9x - 18 by x^2 + 9 ?
so it would be x^2 - x + 7 - x - 2 ?
combine like terms, and it would be x^2 - 2x + 5 , right?
Hmm.. Looks like you have 2 real zeros and 2 imaginary.
Do you still need help?
yes, I don't know what to do next
I got it! Thanks!

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