To determine which of the given trinomials is a perfect square of a binomial, let’s analyze each trinomial step-by-step.
A trinomial in the form [tex]\( ax^2 + bx + c \)[/tex] is a perfect square of a binomial if it can be written as [tex]\( (dx + e)^2 \)[/tex], where [tex]\( dx^2 \)[/tex] results in [tex]\( a \)[/tex], [tex]\( 2de \)[/tex] results in [tex]\( b \)[/tex], and [tex]\( e^2 \)[/tex] results in [tex]\( c \)[/tex].
Let's check each given trinomial:
1. Trinomial: [tex]\( -2x^2 + 12x - 3 \)[/tex]
- To be a perfect square, it would have to match the form [tex]\( (dx + e)^2 \)[/tex].
- Trying different factorization approaches, it’s clear that this trinomial cannot be factored into a perfect square.
2. Trinomial: [tex]\( 16x^2 + 8x + 1 \)[/tex]
- Let's try to factor this as [tex]\( (4x + 1)^2 \)[/tex]:
[tex]\[
(4x + 1)^2 = 16x^2 + 8x + 1
\][/tex]
- This matches exactly with the original trinomial.
- Therefore, [tex]\( 16x^2 + 8x + 1 \)[/tex] is indeed a perfect square trinomial.
3. Trinomial: [tex]\( 9x^2 - 6x + 4 \)[/tex]
- To be a perfect square, it would have to match the form [tex]\( (3x - 2)^2 \)[/tex]:
[tex]\[
(3x - 2)^2 = 9x^2 - 12x + 4
\][/tex]
- This trinomial does not match as the middle term differs.
- Thus, [tex]\( 9x^2 - 6x + 4 \)[/tex] is not a perfect square trinomial.
4. Trinomial: [tex]\( x^2 + 9x + 9 \)[/tex]
- Let's try to factor this as [tex]\( (x + \frac{9}{2})^2 \)[/tex]:
[tex]\[
(x + \frac{9}{2})^2 = x^2 + 9x + \frac{81}{4}
\][/tex]
- This trinomial does not form a perfect square trinomial.
- Therefore, [tex]\( x^2 + 9x + 9 \)[/tex] is not a perfect square trinomial.
5. Trinomial: [tex]\( 9x^2 - 48x + 16 \)[/tex]
- Let's try to factor this as [tex]\( (3x - 4)^2 \)[/tex]:
[tex]\[
(3x - 4)^2 = 9x^2 - 24x + 16
\][/tex]
- Comparing this with the original trinomial, the middle term does not match.
- Thus, [tex]\( 9x^2 - 48x + 16 \)[/tex] is not a perfect square trinomial.
Given this analysis, the trinomial [tex]\( 16x^2 + 8x + 1 \)[/tex] is the only one that is a perfect square of a binomial.
Therefore, the correct answer is:
[tex]\[ 16x^2 + 8x + 1 \][/tex]
