# Polynomial and Rational Functions

• May 5th 2009, 06:28 PM
MathMack
Polynomial and Rational Functions
Hello, I have a few questions that I'm stuck on, not sure what to do.

First one says to: Form a polynomial f(x) with real coefficients having tghe given degree and zeros.

1.) Degree 4; zeros: 3 +2i; 4, Multiplicity 2

The next one says to find the complex zeros of each polynomial function. Write F in factored form.

2.) f(x) = x^3 - 8x^2 + 25x -26

Thanks alot everyone!!
• May 5th 2009, 07:02 PM
HallsofIvy
[QUOTE=MathMack;311940]Hello, I have a few questions that I'm stuck on, not sure what to do.

First one says to: Form a polynomial f(x) with real coefficients having tghe given degree and zeros.

1.) Degree 4; zeros: 3 +2i; 4, Multiplicity 2[quote]
In order to have real coefficients, the 3+ 2i root must be paired with 3- 2i so you want roots 3+ 2i, 3- 2i, 2, 2. The polynomial is (x- (3+ 2i))(x-(3-2i))(x- 2)(x- 2)= (x- 3-2i)(x-3+2i)(x-2)(x-2). Multiply that out.

[quote]The next one says to find the complex zeros of each polynomial function. Write F in factored form.

2.) f(x) = x^3 - 8x^2 + 25x -26
Quote:

I notice that $2^3- 8(2^2)+ 25(2)- 26= 8- 32+ 50- 26= 58- 58= 0$ so x= 2 is one root. Dividing by x- 2, we get $x^2- 6x+ 13$. Now complete the square. Write $x^2- 6x+ 13= x^2- 6x+ 9- 9+ 13= (x- 3)^2- 4= 0$ so $(x- 3)^2= -4$, $x- 3= \pm 2i$, $x= 3\pm 2i$. The roots are x= 2, x= 3+ 2i, and x= 3- 2i. The factored form is (x- 2)(x- 3- 2i)(x- 3+ 2i).
(x- 3
Thanks alot everyone!!
• May 5th 2009, 07:03 PM
Prove It
Quote:

Originally Posted by MathMack
Hello, I have a few questions that I'm stuck on, not sure what to do.

First one says to: Form a polynomial f(x) with real coefficients having tghe given degree and zeros.

1.) Degree 4; zeros: 3 +2i; 4, Multiplicity 2

The next one says to find the complex zeros of each polynomial function. Write F in factored form.

2.) f(x) = x^3 - 8x^2 + 25x -26

Thanks alot everyone!!

What do you mean by "Multiplicity 2"?