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

x=i√14,−i√14,−3+2i,−3−2i

Explanation:

Set the function equal to 0 and solve for x.

Answer 2

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

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)=(-b\pm √(b^2-4ac))/(2a)`

`\mathrm{For\:}\quad a=1,\:b=6,\:c=13:\quad x_(1,\:2)=(-6\pm √(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`