Chapter 4.3, Problem 18AYU

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 18AYU

$\text{The polynomial function }f\left(x\right)=\left(x^4-2x^3+6x^2-2x+5\right)$.

Explanation of Solution

Given:

Degree $=4$, zeros $=i,1+2i$.

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

Since $i,1+2i$ are 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 $i,1+2i,-i,1-2i$ are the complex zeros of the polynomial function $f\left(x\right)$.

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

The factored form $=f\left(x\right)=a\left[\left(x–i\right)\left(x–\left(-i\right)\right]\right[x-\left(1+2i\right)\left]\right[x-\left(1-2i\right)]$.

Let the leading coefficient $a=1$.

Therefore $f\left(x\right)=\left[\left(x–i\right)\left(x+i\right)\right]\left[x-\left(1+2i\right)\right]\left[x-\left(1-2i\right)\right]$.

$=\left(x^2-i^2\right)\left[\left(x-1\right)-2i\right]\left[\left(x-1\right)+2i\right]$

$=\left(x^2+1\right)\left[{\left(x-1\right)}^2-4i^2\right]$

$i^2=-1$

$=\left(x^2+1\right)\left(x^2-2x+1+4\right)$

$=\left(x^2+1\right)\left(x^2-2x+5\right)$

$=\left(x^4-2x^3+6x^2-2x+5\right)$

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

$Y\text{–Intercept }=f\left(x\right)=0=5$