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Which of the following are perfect square trinomials? Choose all that apply.

A. [tex]x^2 + 20x + 100[/tex]

B. [tex]x^2 - 10x + 5[/tex]

C. [tex]x^2 + 6x + 9[/tex]

D. [tex]x^2 + 4x + 4[/tex]

E. [tex]x^2 + 6x - 9[/tex]

F. [tex]x^2 - 8x - 16[/tex]


Sagot :

To determine which of the given trinomials are perfect squares, we need to recognize whether each trinomial can be expressed in the form [tex]\((ax + b)^2 = a^2x^2 + 2abx + b^2\)[/tex].

Let's analyze each trinomial step-by-step.

1. Trinomial: [tex]\(x^2 + 20x + 100\)[/tex]
- This can be expressed as [tex]\((x + 10)^2\)[/tex] because:
[tex]\[ (x + 10)^2 = x^2 + 2 \cdot 10 \cdot x + 10^2 = x^2 + 20x + 100 \][/tex]
Therefore, [tex]\(x^2 + 20x + 100\)[/tex] is a perfect square trinomial.

2. Trinomial: [tex]\(x^2 - 10x + 5\)[/tex]
- This does not match the form [tex]\((ax + b)^2\)[/tex] because the middle term does not equal [tex]\(2ab\)[/tex] (where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are constants and [tex]\( \frac{-10}{2a} = b \)[/tex].

3. Trinomial: [tex]\(x^2 + 6x + 9\)[/tex]
- This can be expressed as [tex]\((x + 3)^2\)[/tex] because:
[tex]\[ (x + 3)^2 = x^2 + 2 \cdot 3 \cdot x + 3^2 = x^2 + 6x + 9 \][/tex]
Therefore, [tex]\(x^2 + 6x + 9\)[/tex] is a perfect square trinomial.

4. Trinomial: [tex]\(x^2 + 4x + 4\)[/tex]
- This can be expressed as [tex]\((x + 2)^2\)[/tex] because:
[tex]\[ (x + 2)^2 = x^2 + 2 \cdot 2 \cdot x + 2^2 = x^2 + 4x + 4 \][/tex]
Therefore, [tex]\(x^2 + 4x + 4\)[/tex] is a perfect square trinomial.

5. Trinomial: [tex]\(x^2 + 6x - 9\)[/tex]
- This doesn't match the form [tex]\((ax + b)^2\)[/tex] for any constants [tex]\(a\)[/tex] and [tex]\(b\)[/tex].

6. Trinomial: [tex]\(x^2 - 8x - 16\)[/tex]
- This doesn't match the form [tex]\((ax + b)^2\)[/tex] for any constants [tex]\(a\)[/tex] and [tex]\(b\)[/tex].

So, the perfect square trinomials from the given list are:

- [tex]\(x^2 + 20x + 100\)[/tex]
- [tex]\(x^2 + 6x + 9\)[/tex]
- [tex]\(x^2 + 4x + 4\)[/tex]

These are the trinomials that can be expressed as the square of a binomial.