Question

Asked Feb 24, 2019

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Find a fourth-degree polynomial function f(x) with real

coeffi cients that has -1, 1, and i as zeros and such that

f(3) = 160.

Step 1

Since two zeros are -1 and 1, so two factors are (x+1) and (x-1), The complex zeros occur in conjugate pairs,so if i is its zero ,then -i is also its zero

Therefore, two other factors are (x+i) and (x-i), that is (x+i) (x-i)=(x^2+1) is also its factor .

So, we assume the polymial as f(x) =a(x+1)(x-1)(x^2+1) as shown on board.

Step 2

Now, we use the condition f(3)=160 and find the value of a

Substitute...

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