Question

Which of the following is the complete list of roots for polynomial function f(x)=(x^2+6x+8)(x^2+6x+13)?

Answer 1

The complete roots of the function $f\left( x \right)= \left({{x^2} + 6x + 8}\right)\left( {{x^2} + 6x + 13}\right)$ are $\boxed{ - 2, - 4, - 3 + 2i, - 3 - 2i}$.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.

$f\left( x \right)=a{x^n} + b{x^{n - 1}}+\ldots + cx + d$

The polynomial function has $n$ roots or zeroes.

Given:

The function is $f\left( x \right)= \left( {{x^2} + 6x + 8}\right)\left( {{x^2} + 6x + 13}\right).$

Explanation:

To obtain the roots of the polynomial function substitute $0$ for $f\left( x \right)$

$= \left({{x^2}+ 6x + 8}\right)\left( {{x^2} + 6x + 13}\right).$

$\left( {{x^2} + 6x + 8}\right)\left( {{x^2} + 6x + 13} \right)= 0$

Either the value of ${x^2} + 6x + 8 = 0$ or ${x^2} + 6x + 13 = 0.$

Solve the function ${x^2} + 6x + 8 = 0$ to obtain the zeroes.

$\begin{aligned}{x^2} + 6x + 8&=0\\{x^2} + 4x + 2x + 8&= 0\\x\left({x + 4}\right) +2\left({x + 4}\right)&= 0\\\left({x + 4}\right)\left({x + 2} \right)&= 0\\x + 4&= 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{or}}\,\,\,\,\,\,\,\,\,\,\,\,x + 2&= 0\\ x=- 4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{or}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x &=- 2 \\\end{aligned}$

Solve the function ${x^2} + 6x + 13 = 0$ to obtain the zeroes.

$\begin{aligned}D&= {b^2} - 4ac\\&= {\left(6 \right)^2} - 4 \times 1\times 13\\&= 36 - 52\\&=- 16\\\end{aligned}$

The roots can be obtained as follows,

$\begin{aligned}x &= \frac{{ - b \pm \sqrt D }}{{2a}}\\x&= \frac{{- 6 \pm\sqrt{ - 16} }}{2}\\x &= \frac{{ - 6 \pm 4\sqrt {- 1} }}{2}\\x&=- 3 \pm 2i \\x&= - 3 + 2i\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{or}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x&=- 3 - 2i\\\end{aligned}$

The complete roots of the function $f\left( x \right)=\left( {{x^2} + 6x + 8} \right)\left({{x^2} + 6x + 13}\right)$ are $\boxed{ - 2, - 4, - 3 + 2i, - 3 - 2i}.$ 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 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.

Answer 2

Equaling the function to zero we have:

$(x ^ 2 + 6x + 8) (x ^ 2 + 6x + 13) = 0$

For the first parenthesis we have:

$(x ^ 2 + 6x + 8) = 0\\(x + 4) (x + 2) = 0$

Therefore the roots are:

$x = - 4\\x = - 2$

For the second parenthesis we have:

$(x ^ 2 + 6x + 13) = 0$

By completing squares we have:

$x ^ 2 + 6x = -13$

$x ^ 2 + 6x + (\frac{6}{2}) ^ 2 = -13 + (\frac{6}{2}) ^ 2\\x ^ 2 + 6x + 9 = -13 + 9\\(x + 3) ^ 2 = - 4\\x + 3 = +/- \sqrt{-4}$

Therefore the roots are:

$x = -3 + 2i\\x = -3 - 2i$

Answer:

The following is the complete list of roots for polynomial function:

C) -2, -4, -3 + 2i, -3-2i