Chapter 2.5, Problem 65E

To determine

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

Answer to Problem 65E

The function as product of linear factors: $\boxed{\boxed{h\left(x\right)=\left(x-1-4i\right)\left(x-1+4i\right)}}$

The zeros of the function are: $\boxed{\boxed{x=1+4i;1-4i}}$

Explanation of Solution

Given information:

  $h\left(x\right)=x^2-2x+17$

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 $n$ linearfactors $f\left(x\right)=a_n\left(x-c_1\right)\left(x-c_2\right)...\left(x-c_n\right)$ 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 alsoa zero of the function.

Calculation:

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

  $x^2-2x+17=0$

We can solve this equation using completing the square method by subtracting17 on both sides of the equation

  $x^2-2x+17-17=0-17$

Simplify on both sides of the equation

  $x^2-2x=-17$

In order to make the left side expression as perfect square trinomial, we need to add square of half of coefficient of x on both sides

  $x^2-2x+1=-17+1$

Simplify on right side of the equation

  $x^2-2x+1=-16$

Write the left side as a perfect square

  ${\left(x-1\right)}^2=-16$

Take square root on both sides

  $\sqrt{{\left(x-1\right)}^2}=\sqrt{-16}$

Split the number in the right side of the equation

  $\sqrt{{\left(x-1\right)}^2}=\sqrt{-1}\cdot \sqrt{16}$

Simplifying square root on both sides of the equation

  $x-1=\sqrt{-1}\cdot \pm 4$

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

  $x-1=\pm 4i$

Add 1 on both sides of the equation

  $x-1+1=1\pm 4i$

Simplify in left side of the equation

  $x=1\pm 4i$

List the zeros of the functions given:

  $x=1+4i;x=1-4i$

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

So $x-\left(1+4i\right)$ & $x-\left(1-4i\right)$ are the factors of the function

Write the function $h\left(x\right)=x^2-2x+17$ as product of linear factors:

  $h\left(x\right)=\left(x-1-4i\right)\left(x-1+4i\right)$

Conclusion:

The zeros of the given function $h\left(x\right)=x^2-2x+17$ are: $x=1+4i;1-4i$

The function written as product of linear factors: $h\left(x\right)=\left(x-1-4i\right)\left(x-1+4i\right)$