SOLUTION: Form a polynomial f(x) with real coefficients having the given degree and zeros. Degree: 18; Zeros: 3 and 1+i. I know that the other zero has to be 1-i, however, I seem to

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Form a polynomial f(x) with real coefficients having the given degree and zeros. Degree: 18; Zeros: 3 and 1+i. I know that the other zero has to be 1-i, however, I seem to       Log On


   

Question 150527: Form a polynomial f(x) with real coefficients having the given degree and zeros.
Degree: 18; Zeros: 3 and 1+i. I know that the other zero has to be 1-i, however, I seem to have misplaced my notes from class and my textbook is not offering much help. Any assistance would be greatly appreciated. Thanks.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Are you sure that the polynomial is not of degree 3? If it's degree 18, then what is the multiplicity of each root?


So I'm going to assume that the polynomial is of degree 3.


Since 3, 1%2Bi, and 1-i are given zeros this means that:


x=3, x=1%2Bi, and x=1-i


Get all terms to the left side in each case


x-3=0, x-%281%2Bi%29=0, and x-%281-i%29=0



%28x-3%29%28x-%281%2Bi%29%29%28x-%281-i%29%29=0 Now use the zero product property in reverse to join the factors.


%28x-3%29%28%28x-1%29%2Bi%29%28%28x-1%29-i%29=0 Regroup the terms


%28x-3%29%28%28x-1%29%5E2-i%5E2%29=0 Factor %28%28x-1%29%2Bi%29%28%28x-1%29-i%29 to get %28x-1%29%5E2-i%5E2 by use of the difference of squares. Note: let A=x-1 and rewrite the problem into %28A%2Bi%29%28A-i%29


%28x-3%29%28%28x-1%29%5E2-%28-1%29%29=0 Rewrite i%5E2 as -1


%28x-3%29%28%28x-1%29%5E2%2B1%29=0 Rewrite %28x-1%29%5E2-%28-1%29 as %28x-1%29%5E2%2B1


x%28x%5E2-2x%2B2%29-3%28x%5E2-2x%2B2%29 Expand. Remember, %28a%2Bb%29%28c%2Bd%2Be%29=a%28c%2Bd%2Be%29%2Bb%28c%2Bd%2Be%29


Distribute.


x%5E3-2%2Ax%5E2%2B2%2Ax-3%2Ax%5E2%2B6%2Ax-6 Multiply.


x%5E3-5%2Ax%5E2%2B8%2Ax-6 Now combine like terms.



So the polynomial of degree 3 that has the roots 3, 1%2Bi, and 1-i is x%5E3-5%2Ax%5E2%2B8%2Ax-6



Notice how if we graph y=x%5E3-5%2Ax%5E2%2B8%2Ax-6, we can visually verify our answer (note: in this case, we can only verify the root 3)

+graph%28+500%2C+500%2C+-10%2C+10%2C+-10%2C+10%2C+x%5E3-5%2Ax%5E2%2B8%2Ax-6+%29+ Graph of y=x%5E3-5%2Ax%5E2%2B8%2Ax-6 with root of x=3