Question

What are all of the real roots of the following polynomial? f(x) = x4 - 13x2 + 36

-3, -2, 2, and 3
-3 and 3
-2 and 2
-3, -1, 1, and 3

Answer 1

x^4 - 13x^2 + 36 = 0

(x^2 - 9)(x^2 - 4) = 0

x^2 - 9 = 0 gives x = 3, -3

x^2 - 4 = 0, gives x = 2, -2.

Answer is A.

Answer 2

ANSWER

A. -3, -2, 2, and 3

Step-by-step explanation

The given polynomial function is,


`f(x) = {x}^(4) - 13 {x}^(2) + 36`

To find the real roots, we equate the function to zero to obtain;


`{x}^(4) - 13 {x}^(2) + 36 = 0`

We can solve this equation as a quadratic equation in

`{x}^(2).`

Thus we rewrite the equation as,


`({x}^(2))^(2) - 13 {x}^(2) + 36 = 0`

We split the middle term to get,


`({x}^(2))^(2) - 9 {x}^(2) - 4 {x}^(2) + 36 = 0.`

We factor to get,


`{x}^(2) ( {x}^(2) - 9) - 4( {x}^(2) - 9) = 0`

We factor to get,


`( {x}^(2) - 9)( {x}^(2) - 4) = 0`

Either


`{x}^(2) - 9 = 0`
or


`{x}^(2) - 4 = 0`


`x = \pm √(9) \: or \: x = \pm √(4)`

`x = \pm3 \: or \: x = \pm2`


`x=-3,-2,2,3`

The correct answer is A