Q. 393 - Intermediate Algebra

Question

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

$169n^3- n$

Step-by-Step Solution

Answer

The Factored form of given polynomial is,

$n(13n-1)(13n+1)$

Step 1. Given Information

We are given a polynomial, 

$169n^3- n$

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

Factoring out greatest common factor $n$, we get

$169n^3-n=n(169n^2-1)$

$169n^3-n=n({\left(13n\right)}^2-1^2)$

Using $a^2-b^2=(a+b)(a-b)$, we get 

$169n^3-n=n(13n-1)(13n+1)$

Step 3. Checking the factorization by multiplying

 Multiplying the factors, we get

$$\begin{gathered} n(13n-1)(13n+1)=169n^3-n \\ n(169n^2-13n+13n-1)=169n^3-n \\ 169n^3-n\times 1=169n^3-n \\ 169n^3-n=169n^3-n \\ LHS=RHS \end{gathered}$$

Hence the factorization is correct.