Q 187 - 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 complete factor of ploynomial is $\left(z-1\right)\left(z+1\right)\left(z^2+1\right)$.

Step 1. Given information

Consider the given ploynomial

$z^4-1$

Step 2. Formula used

$a^2-b^2=\left(a+b\right)\left(a-b\right)$

Step 3. Write the polynomial as squares.

Write the polynomial as perfect squared form.

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

Write them as product of conjugate. 

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