Question

Which will result in a difference of squares? (–7x + 4)(–7x + 4) (–7x + 4)(4 – 7x) (–7x + 4)(–7x – 4) (–7x + 4)(7x – 4)

Answer 1

You can simply expand each product and see whether it gives you a difference of squares.


•  $\mathsf{(-7x+4)\cdot (-7x+4)}$

That's actually  $\mathsf{(-7x+4)^2:}$

     $\mathsf{(-7x+4)^2}\\\\ \mathsf{=(-7x+4)\cdot (-7x+4)}\\\\ \mathsf{=(-7x+4)\cdot (-7x)+(-7x+4)\cdot 4}\\\\ \mathsf{=49x^2-28x-28x+16}$

     $\mathsf{=49x^2-56x+16}$        ✖

which is not a difference of squares.

—————

•  $\mathsf{(-7x+4)\cdot (4-7x)}$

     $\mathsf{=(-7x+4)\cdot 4-(-7x+4)\cdot 7x}\\\\ \mathsf{=-28x+16-(-49x^2+28x)}\\\\ \mathsf{=-28x+16+49x^2-28x}$

     $\mathsf{=49x^2-56x+16}$        ✖

which is not a difference of squares.

—————

•  $\mathsf{(-7x+4)\cdot (-7x-4)}$

     $\mathsf{=(-7x+4)\cdot (-7x)-(-7x+4)\cdot 4}\\\\ \mathsf{=49x^2-28x-(-28x+16)}\\\\ \mathsf{=49x^2-\diagup\!\!\!\!\! 28x+\diagup\!\!\!\!\! 28x-16}\\\\ \mathsf{=49x^2-16}$

     $\mathsf{=(7x)^2-4^2}$        ✔

That is a difference of two squares.

—————

•  $\mathsf{(-7x+4)\cdot (7x-4)}$

     $\mathsf{=(-7x+4)\cdot 7x-(-7x+4)\cdot 4)}\\\\ \mathsf{=-49x^2+28x-(-28x+16)}\\\\ \mathsf{=-49x^2+28x+28x-16}$

     $\mathsf{=-49x^2+56x-16}$        ✖

which is not a difference of squares.

—————

Only the  third option  will result in a difference of squares.


Answer:  (− 7x + 4) · (− 7x − 4).


I hope this helps. =)

Answer 2

Answer:

The expression which will result in difference of two squares is:

(–7x + 4)·(–7x – 4)

Step-by-step explanation:

We know that the formula of the type:

$(a-b).(a+b)=a^2-b^2$

i.e. it is a difference of two square quantities. (a^2 and b^2)

Hence the option which satisfies the following expression is:

(-7x + 4)·(-7x-4)

since,

here $a=-7x$ and $b=4$ and

$(-7x+4).(-7x-4)=(-7x)^2-(4)^2=(7x)^2-4^2$

so the expression is a difference of two square quantities:

$(7x)^2$ and $4^2$

Hence, the correct answer is:

(-7x + 4)·(-7x-4)