Q. 396 - 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 Factored form of given polynomial is,

$(7x+9y)(7x-9y)$

Step 1. Given Information

We are given a polynomial,  

$49x^2-81y^2$

The formula used for factoring using the difference of squares pattern is,

$a^2-b^2=(a+b)(a-b)$

Step 2. Factorizing the polynomial

The given polynomial can be written as,

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

Using $a^2-b^2=(a+b)(a-b)$, we get

$49x^2-81y^2=(7x+9y)(7x-9y)$

Step 3. Checking the factorization by multiplying

 Multiplying the factors, we get 

$$\begin{gathered} (7x+9y)(7x-9y)=49x^2-81y^2 \\ 49x^2-7x\times 9y+9y\times 7x-81y^2=49x^2-81y^2 \\ 49x^2-81y^2=49x^2-81y^2 \\ LHS=RHS \end{gathered}$$

Hence the factorization is correct.