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



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shubhamsrg
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1 fits in.

shubhamsrg
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Got it, break 7 as 92 /

DeadShot
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so \[\pm 1\] is the answer?

shubhamsrg
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nopes.

shubhamsrg
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What do get after writing 7 as 92 ?

DeadShot
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\[x^4  x^3 + 9  2x^2  9x 18\] ?

shubhamsrg
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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 ?

DeadShot
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yeah, factor each binomial, right?

shubhamsrg
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yep.

DeadShot
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(x^2 + 1) (x^2 + 9) for the first one?

shubhamsrg
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try again.

DeadShot
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I'm confused

amoodarya
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x1 = 2
x2 = 1
x3 = 0 + 3i
x4 = 0  3i

DeadShot
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So, substitute (0  3i) and 1 for x^2, making it ((0  3i) + 9(1))?

DeadShot
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so, then it would become (3i  9)

amoodarya
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hint : divide f(x) to x^2+9
you would find your answer!

DeadShot
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So, divide x^4  x^3 + 7x^2  9x  18 by x^2 + 9 ?

DeadShot
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so it would be x^2  x + 7  x  2 ?

DeadShot
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combine like terms, and it would be x^2  2x + 5 , right?

zordoloom
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Hmm.. Looks like you have 2 real zeros and 2 imaginary.

zordoloom
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Do you still need help?

DeadShot
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yes, I don't know what to do next

DeadShot
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I got it! Thanks!