Question

what are the roots of the polynomial function h(x)=3x^2 + 243? (1 point)
the roots are 9 and - 9
the roots are 81i and - 81i
the roots are 9i and - 9i
the roots are 81 and - 81

Explanation:

Step1: Set the polynomial equal to zero.

Set \(h(x)=3x^{2}+243 = 0\).

Step2: Isolate the \(x^{2}\) - term.

First, subtract 243 from both sides: \(3x^{2}=-243\). Then divide both sides by 3: \(x^{2}=- 81\).

Step3: Solve for \(x\).

Take the square - root of both sides. Recall that if \(x^{2}=a\), then \(x=\pm\sqrt{a}\). Here, \(a = - 81\), and \(\sqrt{-81}=\pm9i\) since \(i=\sqrt{-1}\) and \(\sqrt{81} = 9\), so \(x=\pm9i\).

Answer:

The roots are \(9i\) and \(-9i\)