Chapter 2.5, Problem 63E

To determine

To write : the polynomial as the product of linear factors and list all the zeros of function

Answer to Problem 63E

The function as product of linear factors: $\boxed{\boxed{f\left(x\right)=\left(x+6i\right)\left(x-6i\right)}}$

The zeros of the function are: $\boxed{\boxed{x=6i;-6i}}$

Explanation of Solution

Given information:

  $f\left(x\right)=x^2+36$

Concept Involved:

Linear Factorization Theorem: If $f\left(x\right)$ is a polynomial of degree $n$ , where $n>0$ , then $f\left(x\right)$ has precisely $n$ linear factors $f\left(x\right)=a_n\left(x-c_1\right)\left(x-c_2\right)...\left(x-c_n\right)$ where $c_1,c_2,...,c_n$ are complex numbers.

Complex Zeros Occur in Conjugate Pairs: Let f be a polynomial function that has real coefficients. If $a+bi$ , where $b\ne 0$ , is a zero of the function, then the complex conjugates $a-bi$ is also a zero of the function.

Calculation:

To find the zeros of the function we need to set the function to zero $f\left(x\right)=0$

  $x^2+36=0$

Subtract 36 on both sides of the equation

  $x^2+36-36=0-36$

Simplify on both sides of the equation

  $x^2=-36$

Take square root on both sides

  $\sqrt{x^2}=\sqrt{-36}$

Simplifying square root on both sides of the equation

  $\begin{array}{l}x=\sqrt{-1}\cdot \sqrt{36}\\ =\sqrt{-1}\cdot \pm 6\end{array}$

Replace $i$ for $\sqrt{-1}$

  $x=\pm 6i$

List the zeros of the functions given:

  $x=6i;x=-6i$

If $x=k$ is zero, then $x-k$ will be the factor of the function.

So $x+6i$ & $x-6i$ are the factors of the function

Write the function $f\left(x\right)=x^2+36$ as product of linear factors:

  $f\left(x\right)=\left(x+6i\right)\left(x-6i\right)$

Conclusion:

The zeros of the given function $f\left(x\right)=x^2+36$ are: $x=6i;-6i$

The function written as product of linear factors will be of the form $f\left(x\right)=\left(x+6i\right)\left(x-6i\right)$