Question

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

Answer 1

ANSWER

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

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

${x}^{2} - 4 = 0$

$x = \pm \sqrt{9} \: or \: x = \pm \sqrt{4}$
$x = \pm3 \: or \: x = \pm2$

$x=-3,-2,2,3$

The correct answer is A

Answer 2

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.