Chapter A.2, Problem 48E

Solution

To factor the difference of two squares of the equation $16-{\left(x+2\right)}^2$ .

  $-\left(x+6\right)\left(x-2\right)$

Given:

The expression, $16-{\left(x+2\right)}^2$

Concept Used:

• Like terms are the terms that contain the same variable with similar exponents on it. Constants are like terms as well.
• In any expression, terms are separated by plus or minus signs.
• Coefficients are the constants in front of the variable terms.

Calculation:

In order to find the factor of the difference of two squares of the equation $16-{\left(x+2\right)}^2$ first rewrite 16 as $4^2$ that is ;

  $4^2-{\left(x+2\right)}^2$

Since both terms are perfect squares, facter using the difference of square formula,

  $a^2-b^2=\left(a+b\right)\left(a-b\right)$

where $a=4,b=x+2$

  $\begin{array}{l}\left(4+x+2\right)\left(4-\left(x+2\right)\right)\\ \left(x+6\right)\left(-x+2\right)\\ \left(x+6\right)\cdot -1\left(x+2\right)\\ -\left(\left(x+6\right)\left(x-2\right)\right)\\ -\left(x+6\right)\left(x-2\right)\end{array}$

The original expression now become

  $-\left(x+6\right)\left(x-2\right)$