Question

Which of the following is the complete list of roots for the polynomial function f(x)=(x2+6+8)(x2+6x+13)

Answer 1

Answer:

x=i√14,−i√14,−3+2i,−3−2i

Step-by-step explanation:

Set the function equal to  0  and solve for  x.

Answer 2

Answer:

The zeros of given polynomial function $f(x)=(x^2+6x+8)(x^2+6x+13)$ are -4 , -2 $-3+2i,-3-2i$

Step-by-step explanation:

Given polynomial, $f(x)=(x^2+6x+8)(x^2+6x+13)$

We have to find the zeros of the polynomial.

Consider the given polynomial, $f(x)=(x^2+6x+8)(x^2+6x+13)$

Zeros are the point where the value of function is equal to 0  that is

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

Using , zero product property , $a.b=0 \Rightarrow a=0 \ or\ b=0$

we have,

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

We solve it one by one,

consider  $f(x)=(x^2+6x+8)=0$

we solve the quadratic using middle term splitting method,

6x can be written as 4x + 2x ,

$x^2+4x+2x+8=0$

taking x common from first two term and 2 common from last two terms, we have,

$\Rightarrow x(x+4)+2(x+4)=0$

$\Rightarrow (x+4)(x+2)=0$

Again using zero product rule, we have,

$\Rightarrow (x+4)=0$ or  $\Rightarrow (x+2)=0$

$\Rightarrow x=-4$ or  $\Rightarrow x=-2$

Now, consider the second quadratic equation $f(x)=(x^2+6x+13)=0$

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=1,\:b=6,\:c=13:\quad x_{1,\:2}=\frac{-6\pm \sqrt{6^2-4\cdot \:1\cdot \:13}}{2\cdot \:1}$

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

Thus, the zeros of given polynomial function $f(x)=(x^2+6x+8)(x^2+6x+13)$ are -4 , -2 $-3+2i,-3-2i$