Q 6.54 - Intermediate Algebra

Question

Factor: $81y^2-1$.

Step-by-Step Solution

Answer

Factors of $81y^2-1$ are $\left(9y-1\right)\left(9y+1\right)$.

Step 1. Given information.

Given an expression $81y^2-1$.

Step 2. Find if the expression is of the form a 2 - b 2 .

Consider $a=9y,b=1$.

It can be seen that $81y^2-1={\left(9y\right)}^2-1^2$.

Therefore, the expression is of the form .

Step 3. Write the product of conjugates.

Note that .

It follows that $81y^2-1=\left(9y-1\right)\left(9y+1\right)$.

Step 4. Verify.

Multiply the factors as follows.

$\begin{array}{rcl}\left(9y-1\right)\left(9y+1\right)& =& {\left(9y\right)}^2-1\\ & =& 81y^2-1\end{array}$