Q. 397 - Intermediate Algebra

Question

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

$16z^4- 1$

Step-by-Step Solution

Answer

The Factored form of given polynomial is,

$(4z^2+1)(2z+1)(2z-1)$

Step 1. Given Information

We are given a polynomial,

$16z^4- 1$

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.

$16z^4- 1={\left(4z^2\right)}^2-1^2$

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

$16z^4- 1=(4z^2+1)(4z^2-1)$

Now, again using $a^2-b^2=(a+b)(a-b)$, we get

$$\begin{gathered} 16z^4- 1=(4z^2+1)({\left(2z\right)}^2-1^2) \\ 16z^4- 1=(4z^2+1)(2z+1)(2z-1) \end{gathered}$$

Step 3. Checking the factorization by multiplying

Multiplying the factors, we get

$$\begin{gathered} (4z^2+1)(2z+1)(2z-1)=16z^4- 1 \\ (4z^2+1)(4z^2-1)=16z^4- 1 \\ 16z^4-4z^2+4z^2-1=16z^4- 1 \\ 16z^4- 1=16z^4- 1 \\ LHS=RHS \end{gathered}$$

Hence the factorization is correct.