which products result in a difference of squares or a perfect square trinomial? check all that apply. (5x +…

Explanation:

Step1: Recall the formulas

Difference - of - squares: $(a + b)(a - b)=a^{2}-b^{2}$. Perfect - square trinomial: $(a + b)^{2}=a^{2}+2ab + b^{2}$ and $(a - b)^{2}=a^{2}-2ab + b^{2}$.

Step2: Analyze $(5x + 3)(5x - 3)$

Using the difference - of - squares formula with $a = 5x$ and $b = 3$, we get $(5x)^{2}-3^{2}=25x^{2}-9$. So it results in a difference of squares.

Step3: Analyze $(7x + 4)(7x + 4)$

This is in the form $(a + b)^{2}$ with $a = 7x$ and $b = 4$. Expanding, we have $(7x)^{2}+2\times(7x)\times4 + 4^{2}=49x^{2}+56x + 16$, a perfect - square trinomial.

Step4: Analyze $(2x + 1)(x + 2)$

Expanding using FOIL: $2x\times x+2x\times2 + 1\times x+1\times2=2x^{2}+4x+x + 2=2x^{2}+5x + 2$. It is neither a difference of squares nor a perfect - square trinomial.

Step5: Analyze $(4x - 6)(x + 8)$

Expanding using FOIL: $4x\times x+4x\times8-6\times x - 6\times8=4x^{2}+32x-6x - 48=4x^{2}+26x - 48$. It is neither a difference of squares nor a perfect - square trinomial.

Step6: Analyze $(x - 9)(x - 9)$

This is in the form $(a - b)^{2}$ with $a = x$ and $b = 9$. Expanding, we get $x^{2}-2\times x\times9+9^{2}=x^{2}-18x + 81$, a perfect - square trinomial.

Step7: Analyze $(-3x - 6)(-3x + 6)$

Using the difference - of - squares formula with $a=-3x$ and $b = 6$, we have $(-3x)^{2}-6^{2}=9x^{2}-36$, a difference of squares.

Answer:

$(5x + 3)(5x - 3)$, $(7x + 4)(7x + 4)$, $(x - 9)(x - 9)$, $(-3x - 6)(-3x + 6)$