To find the remainder when dividing the polynomial [tex]\(3x^2 + 6x - 4\)[/tex] by the binomial [tex]\(x + 2\)[/tex], we can perform polynomial long division or use the Remainder Theorem. The Remainder Theorem states that the remainder of the division of a polynomial [tex]\(f(x)\)[/tex] by a linear divisor [tex]\(x - c\)[/tex] is equal to [tex]\(f(c)\)[/tex].
Here, we are dividing by [tex]\(x + 2\)[/tex]. We can rewrite the divisor [tex]\(x + 2\)[/tex] as [tex]\(x - (-2)\)[/tex], so in this case, [tex]\(c = -2\)[/tex].
Let's evaluate the polynomial [tex]\(3x^2 + 6x - 4\)[/tex] at [tex]\(x = -2\)[/tex]:
[tex]\[
f(x) = 3x^2 + 6x - 4
\][/tex]
Substitute [tex]\(x = -2\)[/tex]:
[tex]\[
f(-2) = 3(-2)^2 + 6(-2) - 4
\][/tex]
Now calculate step-by-step:
1. Evaluate the squared term:
[tex]\[
(-2)^2 = 4
\][/tex]
2. Multiply by 3:
[tex]\[
3 \times 4 = 12
\][/tex]
3. Evaluate the linear term:
[tex]\[
6 \times (-2) = -12
\][/tex]
4. Combine the evaluated terms and the constant term:
[tex]\[
12 - 12 - 4 = -4
\][/tex]
Thus, the remainder when [tex]\(3x^2 + 6x - 4\)[/tex] is divided by [tex]\(x + 2\)[/tex] is [tex]\(-4\)[/tex].
So the correct answer is:
[tex]\[
\boxed{-4}
\][/tex]
