Question 34bd6

Oct 22, 2017

${\left(3 x\right)}^{4} \left(9 {x}^{2} + 625\right)$

Can also be written as

(3x)^4)*((3x)^2 + 5^4)#

Explanation:

${\left(- 3 x\right)}^{6} + {\left(15 x\right)}^{4}$

$\left({\left(- 3\right)}^{6} \cdot {x}^{6}\right) + \left({\left(15\right)}^{4} \cdot {x}^{4}\right)$

$\left({3}^{6} \cdot {x}^{6}\right) + \left({3}^{4} \cdot {5}^{4} \cdot {x}^{4}\right) a s {\left(- 3\right)}^{6} = {3}^{6}$

Taking common terms out,
$= \left({3}^{4} \cdot {x}^{4}\right) \cdot \left(\left({3}^{2} \cdot {x}^{2}\right) + {5}^{4}\right)$

$= {\left(3 x\right)}^{4} \left(9 {x}^{2} + 625\right)$

Oct 22, 2017

$81 {x}^{4} \left(9 {x}^{2} + 625\right)$

Explanation:

First lets try a test using easy numbers to see if the numbers behave as hoped:

Consider the test
${\left(4 x\right)}^{2} = 16 {x}^{2} \to {\left(2 x \times 2\right)}^{2} = {\left(2 x\right)}^{2} \times {2}^{2} = 4 {x}^{2} \times 4 = 16 {x}^{2}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Answering the question}}$

Using the above approach lets 'force' common factors to occur.
Note that $\left(- 3\right) \times \left(- 5\right) = + 15$

Given: ${\left(- 3 x\right)}^{6} + {\left(15 x\right)}^{4}$

write as: ${\left[- 3 x\right]}^{6} + {\left[- 3 x \times \left(- 5\right)\right]}^{4}$

This is the same as:

$\textcolor{w h i t e}{\text{d}} \left[{\left(- 3 x\right)}^{4} {\left(- 3 x\right)}^{2}\right] + \left[{\left(- 3 x\right)}^{4} {\left(- 5\right)}^{4}\right]$

Factoring out ${\left(- 3 x\right)}^{4} \left[{\left(- 3 x\right)}^{2} + {\left(- 5\right)}^{4}\right]$

Note that ${\left(- 3 x\right)}^{n}$ where $n$ is even gives a positive value
So ${\left(- 3 x\right)}^{4} = {\left(+ 3 x\right)}^{4}$. Also ${\left(- 5\right)}^{4}$ is even.

${\left(3 x\right)}^{4} \left[9 {x}^{2} + {5}^{4}\right]$

$81 {x}^{4} \left(9 {x}^{2} + 625\right)$