Question #34bd6

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Question #34bd6

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Answer 1 · 2017-10-22T19:14:32

$(3x)^4(9x^2 + 625)$

Can also be written as

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

Explanation:

$(-3x)^6 + (15x)^4$

$( (-3)^6 * x^6) + ((15)^4 * x^4)$

$(3^6 * x^6) + (3^4 * 5^4 * x^4) as (-3)^6 = 3^6$

Taking common terms out,
$= (3^4* x^4) * ((3^2 * x^2) + 5^4)$

$= (3x)^4 (9x^2 + 625)$

Answer 2 · 2017-10-22T19:55:11

$81x^4(9x^2+625)$

Explanation:

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

Consider the test
$(4x)^2 =16x^2-> (2x xx2)^2 = (2x)^2xx2^2 = 4x^2xx4=16x^2$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$color(blue)("Answering the question")$

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

Given: $(-3x)^6+(15x)^4$

write as: $[-3x]^6+[-3x xx(-5)]^4$

This is the same as:

$color(white)("d")[(-3x)^4(-3x)^2] + [(-3x)^4(-5)^4]$

Factoring out $(-3x)^4[(-3x)^2+(-5)^4]$

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

$(3x)^4[9x^2+ 5^4]$

$81x^4(9x^2+625)$

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Question #34bd6

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