Q 184 - Intermediate Algebra

Question

In the following exercises, factor completely using the difference of squares pattern, if possible.

$49x^2-81y^2$

Step-by-Step Solution

Answer

The complete factor of trinomial is $\left(7x-9y\right)\left(7x+9y\right)$.

Step 1. Given information

Consider the given trinomial

$49x^2-81y^2$

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. 

$49x^2-81y^2={\left(7x\right)}^2-{\left(9y\right)}^2$

Write them as product of conjugate. 

${\left(7x\right)}^2-{\left(9y\right)}^2=\left(7x-9y\right)\left(7x+9y\right)$