To determine which polynomial has [tex]\(x + 2b\)[/tex] as a factor, we need to factorize each polynomial and see if [tex]\(x + 2b\)[/tex] appears as one of the factors.
### Step-by-Step Factorization:
Option A: [tex]\(3x^2 + 9x + 18b\)[/tex]
We want to check if [tex]\(x + 2b\)[/tex] is a factor.
For [tex]\(x + 2b\)[/tex] to be a factor of [tex]\(3x^2 + 9x + 18b\)[/tex]:
[tex]\[
P(x) = 3x^2 + 9x + 18b
\][/tex]
Perform polynomial division to see if [tex]\(x + 2b\)[/tex] into [tex]\(3x^2 + 9x + 18b\)[/tex].
### Option B: [tex]\(3x^2 + 24x + 18b\)[/tex]
Check if [tex]\(x + 2b\)[/tex] is a factor:
[tex]\[
Q(x) = 3x^2 + 24x + 18b
\][/tex]
### Option C: [tex]\(3x^2 + 30x + 18b\)[/tex]
Check if [tex]\(x + 2b\)[/tex] is a factor:
[tex]\[
R(x) = 3x^2 + 30x + 18b
\][/tex]
### Option D: [tex]\(3x^2 + 39x + 18b\)[/tex]
Check if [tex]\(x + 2b\)[/tex] is a factor:
[tex]\[
S(x) = 3x^2 + 39x + 18b
\][/tex]
### Let's derive for each polynomial:
### For [tex]\(P(x) = 3x^2 + 9x + 18b\)[/tex],
If [tex]\(x + 2b\)[/tex] is a factor, then by synthetic division:
[tex]\(P(-2b) = 0\)[/tex]
Substitute [tex]\(x = -2b\)[/tex] in [tex]\(P(x)\)[/tex]:
[tex]\[
3(-2b)^2 + 9(-2b) + 18b = 12b^2 - 18b + 18b = 12b^2
\][/tex]
Since [tex]\(12b^2 \neq 0\)[/tex], [tex]\(x + 2b\)[/tex] is not a factor.
### For [tex]\(Q(x) = 3x^2 + 24x + 18b\)[/tex],
If [tex]\(x + 2b\)[/tex] is a factor, then by synthetic division:
[tex]\(Q(-2b) = 0\)[/tex]
Substitute [tex]\(x = -2b\)[/tex] in [tex]\(Q(x)\)[/tex]:
[tex]\[
3(-2b)^2 + 24(-2b) + 18b = 12b^2 - 48b + 18b = 12b^2 - 30b
\][/tex]
Since [tex]\( 12b(b - 2.5)\)[/tex], [tex]\(x + 2b\)[/tex] is not a factor.
### For [tex]\(R(x) = 3x^2 + 30x + 18b\)[/tex],
If [tex]\(x + 2b\)[/tex] is a factor, then by synthetic division:
[tex]\(R(-2b) = 0\)[/tex]
Substitute [tex]\(x = -2b\)[/tex] in [tex]\(R(x)\)[/tex]:
[tex]\[
3(-2b)^2 + 30(-2b) + 18b = 12b^2 - 60b + 18b = 12b^2 - 42b
\][/tex]
Since [tex]\( 12b(b - 3.5)\)[/tex], [tex]\(x + 2b\)[/tex] is not a factor.
### For [tex]\(S(x) = 3x^2 + 39x + 18b\)[/tex],
If [tex]\(x + 2b\)[/tex] is a factor, then by synthetic division:
[tex]\(S(-2b) = 0\)[/tex]
Substitute [tex]\(x = -2b\)[/tex] in [tex]\(S(x)\)[/tex]:
[tex]\[
3(-2b)^2 + 39(-2b) + 18b = 12b^2 - 78b + 18b = 12b^2 - 60b
\][/tex]
Since [tex]\( 12b(b - 5b)\)[/tex], [tex]\(x + 2b\)[/tex].
### The right polynomial is
[tex]\[
\boxed{3}
\][/tex]
