Chapter 1.3, Problem 5E

To determine

To fill: The blank in the statement “The Special Factoring Formula for the difference of squares is $A^2-B^2=\underset{\_}{}$. So $4x^2-25$ factor as $\underset{\_}{}$”.

Answer to Problem 5E

The complete statement is “The Special Factoring Formula for the difference of squares is $A^2-B^2=\underset{\_}{\left(A+B\right)\left(A-B\right)}$. So $4x^2-25$ factor as $\underset{\_}{\left(2x+5\right)\left(2x-5\right)}$”.

Explanation of Solution

Formula used:

If A and B are two real numbers then the special product formula for the “difference of squares” is $A^2-B^2=\left(A+B\right)\left(A-B\right)$.

Calculation:

The given expression is $4x^2-25$ which of the form $A^2-B^2$ where $A^2=4x^2,B^2=25$.

Obtain the value of A as shown below.

$\begin{array}{l}\sqrt{A^2}=\sqrt{4x^2}\\ A=2x\end{array}$

Similarly, obtain the value of B as follows.

$\begin{array}{l}\sqrt{B^2}=\sqrt{25}\\ B=5\end{array}$

Substitute $A=2x$ and $B=5$ in $A^2-B^2=\left(A+B\right)\left(A-B\right)$ as follows:

$4x^2-25=\left(2x+5\right)\left(2x-5\right)$

Therefore, the Special Factoring Formula for the difference of squares is $A^2-B^2=\underset{\_}{\left(A+B\right)\left(A-B\right)}$. So $4x^2-25$ factor as $\underset{\_}{\left(2x+5\right)\left(2x-5\right)}$.