Chapter 4.3, Problem 17AYU

In Problems 17-22, form a polynomial function $f\left(x\right)$ with real coefficients having the given degree and zeros. Answers will vary depending on the choice of the leading coefficient. Use a graphing utility to graph the function and verify the result.

Degree 4; zeros: $3+2i$; 4, multiplicity 2

To determine

To form: A polynomial function with real coefficients having the given degree and zeros and to use the graphing utility to graph the function and to verify the result.

Answer to Problem 17AYU

$\text{The polynomial function }f\left(x\right)=\left(x^4-14x^3+77x^2-200x+208\right)$.

Explanation of Solution

Given:

Degree $=4$, zeros $=3+2i,$ 4 multiplicity 2.

Since the degree $=4$ there will be 4 zeros.

Since $3+2i$ is the complex zero of the polynomial function $f\left(x\right)$ and $f\left(x\right)$ has coefficients that are real numbers Complex zero appear as conjugate pairs.

By the conjugate pair theorem $\left(3+2i\right)$ and $\left(3–2i\right)$ are the complex zeros of the polynomial function $f\left(x\right)$.

The real zero $=4$ multiplicity 2.

By factor theorem $f\left(c\right)=0,\left(x-c\right)=\text{ factor of }f\left(x\right)$.

The factored form $=f\left(x\right)=a\left(x–4\right)\left(x–4\right)\left[x-\left(3+2i\right)\right]\left[x-\left(3-2i\right)\right]$.

Let the leading coefficient $a=1$.

Therefore $f\left(x\right)=\left(x–4\right)\left(x–4\right)\left[x-\left(3+2i\right)\right]\left[x-\left(3-2i\right)\right]$.

$=\left(x^2-8x+16\right)\left[\left(x-3\right)+2i\right]\left[\left(x-3\right)-2i\right]$

$=\left(x^2-8x+16\right)\left[{\left(x-3\right)}^2-4i^2\right]$

$i^2=-1$

$=\left(x^2-8x+16\right)\left(x^2-6x+9+4\right)$

$=\left(x^2-8x+16\right)\left(x^2-6x+13\right)$

$=\left(x^4-14x^3+77x^2-200x+208\right)$

Degree $=4$ at most turning points $=3$.

$x\text{-Intercept }=4,Y\text{-intercept }=f\left(x\right)=0=208$