SOLUTION: Form a polynomial f(x) with real coefficients having the given degree and zeros. Degree 5; zeros: -4; -i; -3+i

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Form a polynomial f(x) with real coefficients having the given degree and zeros. Degree 5; zeros: -4; -i; -3+i      Log On


   



Question 1188212: Form a polynomial f(x) with real coefficients having the given degree and zeros.
Degree 5; zeros: -4; -i; -3+i

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
Instead of doing your homework for you, I will do one exactly like yours
step-by-step only with slightly different numbers.  I'll do this one instead:
Form a polynomial f(x) with real coefficients having the given degree and zeros.
Degree 5; zeros: -5; -2i; -4+2i
If a polynomial has a complex imaginary zero a+bi, then a-bi is also a zero.

So all the zeros are: -5; -2i; +2i; -4+2i; -4-2i 

Set x equal to each:

x = -5;  x = -2i;   x = +2i;  x = -4+2i;   x = -4-2i

Get 0 on the right of each:

x+5 = 0; x+2i = 0; x-2i = 0;  x+4-2i = 0;  x+4+2i = 0

Multiply all the left and right sides together:

%28x%2B5%29%28x%2B2i%29%28x-2i%29%28x%2B4-2i%29%28x%2B4%2B2i%29+=+0

%28x%2B5%29%28%28x%2B2i%29%28x-2i%29%5E%22%22%29%28x%2B4-2i%29%28x%2B4%2B2i%29+=+0

%28x%2B5%29%28x%5E2-4i%5E2%29%28x%2B4-2i%29%28x%2B4%2B2i%29+=+0

%28x%2B5%29%28x%5E2-4%28-1%29%29%28x%2B4-2i%29%28x%2B4%2B2i%29+=+0

%28x%2B5%29%28x%5E2%2B4%29%28x%2B4-2i%29%28x%2B4%2B2i%29+=+0

%28x%2B5%29%28x%5E2%2B4%29%28%28x%2B4%29%5E2%5E%22%22-4i%5E2%29+=+0

%28x%2B5%29%28x%5E2%2B4%29%28%28x%2B4%29%5E2%5E%22%22-4%28-1%29%29+=+0

%28x%2B5%29%28x%5E2%2B4%29%28%28x%2B4%29%5E2%2B4%29+=+0

%28x%2B5%29%28x%5E2%2B4%29%28%28x%5E2%2B8x%2B16%29%5E%22%22%2B4%29+=+0

%28x%2B5%29%28x%5E2%2B4%29%28x%5E2%2B8x%2B20%29+=+0

%28x%2B5%29%28x%5E4%2B8x%5E3%2B20x%5E2%2B4x%5E2%2B32x%2B80%29+=+0

%28x%2B5%29%28x%5E4%2B8x%5E3%2B24x%5E2%2B32x%2B80%29+=+0

x%5E5%2B8x%5E4%2B24x%5E3%2B32x%5E2%2B80x%2B5x%5E4%2B40x%5E3%2B120x%5E2%2B160x%2B400=0

x%5E5%2B13x%5E4%2B64x%5E3%2B152x%5E2%2B240x%2B400=0

So a polynomial function of degree 5 which we must set = 0, to find
the zeros: -5; -2i; -4+2i  is the left side of the preceding equation:

%22f%28x%29%22=x%5E5%2B13x%5E4%2B64x%5E3%2B152x%5E2%2B240x%2B400

Edwin