To determine the roots of the polynomial function [tex]\( f(x) = (x^2 + 6x + 8)(x^2 + 6x + 13) \)[/tex], we'll need to find the roots of each quadratic factor separately.
### Step 1: Find the Roots of [tex]\( x^2 + 6x + 8 \)[/tex]
To find the roots of [tex]\( x^2 + 6x + 8 \)[/tex], we can use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 1 \)[/tex], [tex]\( b = 6 \)[/tex], and [tex]\( c = 8 \)[/tex].
Plug these values into the quadratic formula:
[tex]\[ x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot 8}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-6 \pm \sqrt{36 - 32}}{2} \][/tex]
[tex]\[ x = \frac{-6 \pm \sqrt{4}}{2} \][/tex]
[tex]\[ x = \frac{-6 \pm 2}{2} \][/tex]
This gives us two roots:
[tex]\[ x = \frac{-6 + 2}{2} = -2 \][/tex]
[tex]\[ x = \frac{-6 - 2}{2} = -4 \][/tex]
So, the roots of [tex]\( x^2 + 6x + 8 \)[/tex] are [tex]\( x = -2 \)[/tex] and [tex]\( x = -4 \)[/tex].
### Step 2: Find the Roots of [tex]\( x^2 + 6x + 13 \)[/tex]
Next, find the roots of [tex]\( x^2 + 6x + 13 \)[/tex] using the quadratic formula where [tex]\( a = 1 \)[/tex], [tex]\( b = 6 \)[/tex], and [tex]\( c = 13 \)[/tex]:
[tex]\[ x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot 13}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-6 \pm \sqrt{36 - 52}}{2} \][/tex]
[tex]\[ x = \frac{-6 \pm \sqrt{-16}}{2} \][/tex]
Since we have a negative number inside the square root, the roots will be complex numbers:
[tex]\[ x = \frac{-6 \pm 4i}{2} \][/tex]
[tex]\[ x = -3 \pm 2i \][/tex]
So, the roots of [tex]\( x^2 + 6x + 13 \)[/tex] are [tex]\( x = -3 + 2i \)[/tex] and [tex]\( x = -3 - 2i \)[/tex].
### Combine the Roots
Combining the roots from both quadratic factors, we get the complete list of roots for the polynomial function [tex]\( f(x) = (x^2 + 6x + 8)(x^2 + 6x + 13) \)[/tex]:
[tex]\[ x = -2, -4, -3 + 2i, -3 - 2i \][/tex]
Therefore, the correct answer is:
[tex]\[ -2, -4, -3 + 2i, -3 - 2i \][/tex]
