Asked by Cariline
                Find the polynomial function f with real coefficients that has given degree, zeros, and solution point.
Degree: 3
Zeros: -2,1- root of 2i
Solution point: f(-1)=-54
            
        Degree: 3
Zeros: -2,1- root of 2i
Solution point: f(-1)=-54
Answers
                    Answered by
            mathhelper
            
    since complex solutions appear in conjugate pairs, the third solution must be 1 + √2 i
so the function has the form
f(x) = a(x+2)(x - 1 + √2 i)(x - 1 - √2 i)
= a(x-2)(x^2 - 2x + 3)
since f(-1) = a(-3)(1 + 2 + 3) = -18a
so -18a = -54
a = 3
f(x) = 3(x+2)(x^2 - 2x + 3)
    
so the function has the form
f(x) = a(x+2)(x - 1 + √2 i)(x - 1 - √2 i)
= a(x-2)(x^2 - 2x + 3)
since f(-1) = a(-3)(1 + 2 + 3) = -18a
so -18a = -54
a = 3
f(x) = 3(x+2)(x^2 - 2x + 3)
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.