Question

solve the polynomial equation. in order to obtain the first root, use synthetic division to test the possible rational roots.\\(x^3 + 3x^2 - 4x - 12 = 0\\)\\(\bigcirc \\{-3\\}\\)\\(\bigcirc \\{-3, -2, 2\\}\\)\\(\bigcirc \\{-2\\}\\)\\(\bigcirc \\{-2, 2, 3\\}\\)

Explanation:

Step1: List possible rational roots

By Rational Root Theorem, possible roots are $\pm1, \pm2, \pm3, \pm4, \pm6, \pm12$

Step2: Test root $x=-3$ via synthetic division

$$\begin{array}{r|rrrr} -3 & 1 & 3 & -4 & -12 \\ & & -3 & 0 & 12 \\ \hline & 1 & 0 & -4 & 0 \end{array}$$

Since remainder is 0, $x=-3$ is a root. The polynomial factors to $(x+3)(x^2-4)=0$

Step3: Factor the quadratic

$x^2-4=(x-2)(x+2)$, so equation becomes $(x+3)(x-2)(x+2)=0$

Step4: Solve for $x$

Set each factor to 0: $x+3=0 \implies x=-3$; $x-2=0 \implies x=2$; $x+2=0 \implies x=-2$

Answer:

B. {-3, -2, 2}