Question

the fundamental theorem of algebra quick
what are the roots of $f(x) = 6x^2 + 216$? (1 point)
$\pm 6i$
$\pm 36$
$\pm 6$
$\pm 36i$

Explanation:

Step1: Set \( f(x) = 0 \)

To find the roots, we start by setting \( f(x) = 0 \), so we have the equation \( 6x^{2}+216 = 0 \).

Step2: Isolate \( x^{2} \)

First, subtract 216 from both sides: \( 6x^{2}=- 216 \). Then divide both sides by 6: \( x^{2}=\frac{-216}{6}=-36 \).

Step3: Solve for \( x \)

Take the square root of both sides. Remember that \( \sqrt{-a}=i\sqrt{a} \) for \( a>0 \). So \( x = \pm\sqrt{-36}=\pm6i \).

Answer:

\( \pm6i \) (corresponding to the option "±6i")