To find the complete list of roots for the polynomial function [tex]\( f(x) = (x^2 + 6x + 8)(x^2 + 6x + 13) \)[/tex], we need to solve [tex]\( f(x) = 0 \)[/tex].
The polynomial function is factored into two quadratic equations:
[tex]\[ (x^2 + 6x + 8)(x^2 + 6x + 13) = 0 \][/tex]
We will solve each quadratic equation separately:
1. For [tex]\( x^2 + 6x + 8 = 0 \)[/tex]:
This quadratic equation can be factored as:
[tex]\[ x^2 + 6x + 8 = (x + 2)(x + 4) = 0 \][/tex]
So, the roots are:
[tex]\[ x = -2 \quad \text{and} \quad x = -4 \][/tex]
2. For [tex]\( x^2 + 6x + 13 = 0 \)[/tex]:
This quadratic equation does not have real roots, so we use the quadratic formula to find the complex roots. The quadratic formula is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 6 \)[/tex], and [tex]\( c = 13 \)[/tex].
We calculate the discriminant:
[tex]\[ b^2 - 4ac = 6^2 - 4(1)(13) = 36 - 52 = -16 \][/tex]
Since the discriminant is negative, the roots will be complex numbers. Now we find those roots:
[tex]\[ x = \frac{-6 \pm \sqrt{-16}}{2 \cdot 1} = \frac{-6 \pm 4i}{2} = -3 \pm 2i \][/tex]
So, the roots are:
[tex]\[ x = -3 + 2i \quad \text{and} \quad x = -3 - 2i \][/tex]
Combining all the roots from both quadratic equations, we get the complete list of roots:
[tex]\[ -2, -4, -3 + 2i, -3 - 2i \][/tex]
Thus, the correct answer is:
[tex]\[ -2, -4, -3 + 2i, -3 - 2i \][/tex]
Therefore, the complete list of roots for the polynomial function [tex]\( f(x) \)[/tex] is:
[tex]\[ \boxed{-2,-4,-3+2i,-3-2i} \][/tex]
