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Question 1002112: Find a 4th degree polynomial with zeroes of 2,-2,3i and f(0)=-108. Our math book doesn't explain anything on how to solve this, I have searched online and believe I have the solution. I want some verification if i am correct or not.
f(x)=(x-3i)(x+3i)(x+2)(x-2)
(x^2+9)(x^2-4)
(x^4-5x-36)
I am unsure for the rest. and what to do with f(0)=-108
Answer by ikleyn(52778)   (Show Source):
You can put this solution on YOUR website! .
Find a 4th degree polynomial with zeroes of 2,-2,3i and f(0)=-108.
Our math book doesn't explain anything on how to solve this, I have searched online and believe I have the solution.
I want some verification if i am correct or not.
f(x)=(x-3i)(x+3i)(x+2)(x-2)
(x^2+9)(x^2-4)
(x^4-5x-36)
I am unsure for the rest. and what to do with f(0)=-108
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f(x)=(x-3i)(x+3i)(x+2)(x-2) this line is correct.
(x^2+9)(x^2-4) this line is correct.
(x^4-5x-36) this line is incorrect. It should be (x^4+5x^2-36).
So far, you get the polynomial p(x) = x^4+5x^2-36.
Calculate its value at x = 0. Simply substitute x=0 into p(x). You will get
p(0) = 0^4 +5*0^2 - 36 = -36.
Therefore, apply the factor 3 to p(x): f(x) = 3*p(x) to get the polynomial f(x) which has the same zeroes as p(x)
and has the value f(0) = -108 at x=0.
Now your polynomial and your answer is f(x) = 3*(x^4+5x^2-36) = 3x^4 + 15x^2 - 108.
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