Respuesta :
Equaling the function to zero we have:
[tex] (x ^ 2 + 6x + 8) (x ^ 2 + 6x + 13) = 0
[/tex]
For the first parenthesis we have:
[tex] (x ^ 2 + 6x + 8) = 0
(x + 4) (x + 2) = 0
[/tex]
Therefore the roots are:
[tex] x = - 4
x = - 2
[/tex]
For the second parenthesis we have:
[tex] (x ^ 2 + 6x + 13) = 0
[/tex]
By completing squares we have:
[tex] x ^ 2 + 6x = -13
[/tex]
[tex] x ^ 2 + 6x + (\frac{6}{2}) ^ 2 = -13 + (\frac{6}{2}) ^ 2
x ^ 2 + 6x + 9 = -13 + 9
(x + 3) ^ 2 = - 4
x + 3 = +/- \sqrt{-4}
[/tex]
Therefore the roots are:
[tex] x = -3 + 2i
x = -3 - 2i
[/tex]
Answer:
The following is the complete list of roots for polynomial function:
C) -2, -4, -3 + 2i, -3-2i
The complete roots of the function [tex]f\left( x \right)= \left({{x^2} + 6x + 8}\right)\left( {{x^2} + 6x + 13}\right)[/tex] are [tex]\boxed{ - 2, - 4, - 3 + 2i, - 3 - 2i}[/tex].Option (C) is correct.
Further explanation:
The Fundamental Theorem of Algebra states that the polynomial has n roots if the degree of the polynomial is n.
[tex]f\left( x \right)=a{x^n} + b{x^{n - 1}}+\ldots + cx + d[/tex]
The polynomial function has [tex]n[/tex] roots or zeroes.
Given:
The function is [tex]f\left( x \right)= \left( {{x^2} + 6x + 8}\right)\left( {{x^2} + 6x + 13}\right).[/tex]
Explanation:
To obtain the roots of the polynomial function substitute [tex]0[/tex] for [tex]f\left( x \right)[/tex]
[tex]= \left({{x^2}+ 6x + 8}\right)\left( {{x^2} + 6x + 13}\right).[/tex]
[tex]\left( {{x^2} + 6x + 8}\right)\left( {{x^2} + 6x + 13} \right)= 0[/tex]
Either the value of [tex]{x^2} + 6x + 8 = 0[/tex] or [tex]{x^2} + 6x + 13 = 0.[/tex]
Solve the function [tex]{x^2} + 6x + 8 = 0[/tex] to obtain the zeroes.
[tex]\begin{aligned}{x^2} + 6x + 8&=0\\{x^2} + 4x + 2x + 8&= 0\\x\left({x + 4}\right) +2\left({x + 4}\right)&= 0\\\left({x + 4}\right)\left({x + 2} \right)&= 0\\x + 4&= 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{or}}\,\,\,\,\,\,\,\,\,\,\,\,x + 2&= 0\\ x=- 4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{or}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x &=- 2 \\\end{aligned}[/tex]
Solve the function [tex]{x^2} + 6x + 13 = 0[/tex] to obtain the zeroes.
[tex]\begin{aligned}D&= {b^2} - 4ac\\&= {\left(6 \right)^2} - 4 \times 1\times 13\\&= 36 - 52\\&=- 16\\\end{aligned}[/tex]
The roots can be obtained as follows,
[tex]\begin{aligned}x &= \frac{{ - b \pm \sqrt D }}{{2a}}\\x&= \frac{{- 6 \pm\sqrt{ - 16} }}{2}\\x &= \frac{{ - 6 \pm 4\sqrt {- 1} }}{2}\\x&=- 3 \pm 2i \\x&= - 3 + 2i\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{or}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x&=- 3 - 2i\\\end{aligned}[/tex]
The complete roots of the function [tex]f\left( x \right)=\left( {{x^2} + 6x + 8} \right)\left({{x^2} + 6x + 13}\right)[/tex] are [tex]\boxed{ - 2, - 4, - 3 + 2i, - 3 - 2i}.[/tex] Option (C) is correct.
Learn more:
1. Learn more about inverse of the functionhttps://brainly.com/question/1632445.
2. Learn more about equation of circle brainly.com/question/1506955.
3. Learn more about range and domain of the function https://brainly.com/question/3412497
Answer details:
Grade: High School
Subject: Mathematics
Chapter: Linear equation
Keywords: roots,linear equation, quadratic equation, zeros, function, polynomial, solution, cubic function, degree of the function.
