Q 176 - Intermediate Algebra

Question

In the following exercises, factor completely using the difference of squares pattern, if possible.
$169q^2-1$

Step-by-Step Solution

Answer

The complete factor of trinomial is $\left(13q-1\right)\left(13q+1\right)$.

Step 1. Given information

Consider the given trinomial.

$169q^2-1$

Step 2. Formula used

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

Step 3. Write the trinomial as squares.

Write the trinomial as perfect squared form.

$169q^2-1={\left(13q\right)}^2-1^2$

Write them as product of conjugate.

${\left(13q\right)}^2-1^2=\left(13q-1\right)\left(13q+1\right)$