Question

16b2c12–0.25 Factor Completey

Answer 1

Answer:

$(4bc^{6} - \frac{1}{2})(4bc^{6} + \frac{1}{2})$

Step-by-step explanation:

Given

$16b^2c^{12} - 0.25$

Required

Factor Completely

Follow the steps below;

Rewrite 0.25 as a fraction

$16b^2c^{12} - \frac{25}{100}$

Simplify fraction to lowest term

$16b^2c^{12} - \frac{1}{4}$

Expand $16b^2c^{12}$

$4^2b^2c^{6*2} - \frac{1}{4}$

$(4bc^{6})^2 - \frac{1}{4}$

Expand $\frac{1}{4}$

$(4bc^{6})^2 - \frac{1}{2}*\frac{1}{2}$

$(4bc^{6})^2 -( \frac{1}{2})^2$

From laws of product of two squares

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

So,

$(4bc^{6})^2 -( \frac{1}{2})^2$ is equivalent to

$(4bc^{6} - \frac{1}{2})(4bc^{6} + \frac{1}{2})$

The expression cannot be factorized any further;

Hence, the factor of $16b^2c^{12} - 0.25$ is $(4bc^{6} - \frac{1}{2})(4bc^{6} + \frac{1}{2})$