Answer:
[tex](4bc^{6} - \frac{1}{2})(4bc^{6} + \frac{1}{2})[/tex]
Step-by-step explanation:
Given
[tex]16b^2c^{12} - 0.25[/tex]
Required
Factor Completely
Follow the steps below;
Rewrite 0.25 as a fraction
[tex]16b^2c^{12} - \frac{25}{100}[/tex]
Simplify fraction to lowest term
[tex]16b^2c^{12} - \frac{1}{4}[/tex]
Expand [tex]16b^2c^{12}[/tex]
[tex]4^2b^2c^{6*2} - \frac{1}{4}[/tex]
[tex](4bc^{6})^2 - \frac{1}{4}[/tex]
Expand [tex]\frac{1}{4}[/tex]
[tex](4bc^{6})^2 - \frac{1}{2}*\frac{1}{2}[/tex]
[tex](4bc^{6})^2 -( \frac{1}{2})^2[/tex]
From laws of product of two squares
[tex]a^2 - b^2 = (a+b)(a-b)[/tex]
So,
[tex](4bc^{6})^2 -( \frac{1}{2})^2[/tex] is equivalent to
[tex](4bc^{6} - \frac{1}{2})(4bc^{6} + \frac{1}{2})[/tex]
The expression cannot be factorized any further;
Hence, the factor of [tex]16b^2c^{12} - 0.25[/tex] is [tex](4bc^{6} - \frac{1}{2})(4bc^{6} + \frac{1}{2})[/tex]
