To find the remainder when the polynomial [tex]\(x^3 + 3x^2 + 5x + 6\)[/tex] is divided by [tex]\(x + 2\)[/tex], we can use polynomial division. Specifically, we'll employ the Remainder Theorem, which states that the remainder of the division of a polynomial [tex]\(f(x)\)[/tex] by a linear divisor [tex]\(x - a\)[/tex] is [tex]\(f(a)\)[/tex].
In this case, the divisor is [tex]\(x + 2\)[/tex], which can be rewritten as [tex]\(x - (-2)\)[/tex]. Thus, we need to find the value of the polynomial [tex]\(x^3 + 3x^2 + 5x + 6\)[/tex] at [tex]\(x = -2\)[/tex].
Let's evaluate the polynomial at [tex]\(x = -2\)[/tex]:
[tex]\[
f(x) = x^3 + 3x^2 + 5x + 6
\][/tex]
[tex]\[
f(-2) = (-2)^3 + 3(-2)^2 + 5(-2) + 6
\][/tex]
Now compute each term step-by-step:
- [tex]\((-2)^3 = -8\)[/tex]
- [tex]\(3(-2)^2 = 3 \cdot 4 = 12\)[/tex]
- [tex]\(5(-2) = -10\)[/tex]
- Constant term is [tex]\(6\)[/tex]
Combine these values:
[tex]\[
f(-2) = -8 + 12 - 10 + 6
\][/tex]
Simplify the expression:
[tex]\[
f(-2) = -8 + 12 = 4
\][/tex]
[tex]\[
4 - 10 = -6
\][/tex]
[tex]\[
-6 + 6 = 0
\][/tex]
Therefore, the remainder when [tex]\(x^3 + 3x^2 + 5x + 6\)[/tex] is divided by [tex]\(x + 2\)[/tex] is [tex]\(0\)[/tex]. The polynomial division yields no remainder.
