To solve the given polynomial [tex]\( Q(x) = x^4 + 18x^2 + 81 \)[/tex]:
### Step 1: Factor the Polynomial
First, notice that the polynomial can be written as a quadratic expression in terms of [tex]\( x^2 \)[/tex]:
[tex]\[ Q(x) = (x^2)^2 + 18(x^2) + 81 \][/tex]
Let [tex]\( y = x^2 \)[/tex]. Then, the polynomial in terms of [tex]\( y \)[/tex] is:
[tex]\[ Q(y) = y^2 + 18y + 81 \][/tex]
Now, let's factor this quadratic expression. We need two numbers that multiply to 81 and add up to 18. Those numbers are 9 and 9:
[tex]\[ y^2 + 18y + 81 = (y + 9)(y + 9) = (y + 9)^2 \][/tex]
Substituting [tex]\( y = x^2 \)[/tex] back in, we get:
[tex]\[ Q(x) = (x^2 + 9)^2 \][/tex]
### Step 2: Find the Zeros of the Polynomial
We have the factorized form:
[tex]\[ Q(x) = (x^2 + 9)^2 \][/tex]
To find the zeros, set [tex]\( Q(x) = 0 \)[/tex]:
[tex]\[ (x^2 + 9)^2 = 0 \][/tex]
Taking the square root of both sides:
[tex]\[ x^2 + 9 = 0 \][/tex]
Solving for [tex]\( x^2 \)[/tex]:
[tex]\[ x^2 = -9 \][/tex]
Taking the square root of both sides, we get:
[tex]\[ x = \pm \sqrt{-9} = \pm 3i \][/tex]
### Step 3: Determine the Multiplicity of Each Zero
Each zero [tex]\( 3i \)[/tex] and [tex]\(-3i\)[/tex] comes from the factor [tex]\((x^2 + 9)^2\)[/tex], hence each has multiplicity 2.
### Summary
1. The completely factored form of the polynomial is:
[tex]\[
Q(x) = (x^2 + 9)^2
\][/tex]
2. The zeros and their multiplicities are:
[tex]\[
\begin{array}{l}
x = -3i \text{ with multiplicity } 2 \\
x = 3i \text{ with multiplicity } 2
\end{array}
\][/tex]
