To factor the quadratic expression [tex]\(9x^2 + 6x + 1\)[/tex], we need to determine what the expression simplifies to when written in the form of [tex]\((ax + b)^2\)[/tex]. Follow these steps:
1. Recognize the standard form of a quadratic expression:
[tex]\[
ax^2 + bx + c
\][/tex]
Here, [tex]\( a = 9 \)[/tex], [tex]\( b = 6 \)[/tex], and [tex]\( c = 1 \)[/tex].
2. Set up the factored form as [tex]\((dx + e)^2\)[/tex] and expand it:
[tex]\[
(dx + e)^2 = d^2x^2 + 2dex + e^2
\][/tex]
3. Compare coefficients:
- The coefficient of [tex]\(x^2\)[/tex] in [tex]\((dx + e)^2\)[/tex] should match [tex]\(9x^2\)[/tex] from the original expression.
- Hence, [tex]\( d^2 = 9 \)[/tex] which gives [tex]\( d = 3 \)[/tex] (since [tex]\(3^2 = 9\)[/tex]).
- The constant term [tex]\(e^2\)[/tex] should match [tex]\(1\)[/tex].
- Therefore, [tex]\( e^2 = 1 \)[/tex] which gives [tex]\( e = 1 \)[/tex] (since [tex]\(1^2 = 1\)[/tex]).
4. The middle term [tex]\(2dex\)[/tex] of the expanded form should match the middle term [tex]\(6x\)[/tex] of the original expression.
[tex]\[
2(3)(1)x = 6x
\][/tex]
Given these comparisons, the correct factorized form of the quadratic expression [tex]\(9x^2 + 6x + 1\)[/tex] is therefore:
[tex]\[
(3x + 1)^2
\][/tex]
So, the correct answer is:
[tex]\[
(3x + 1)^2
\][/tex]
