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# How would I factor 4x^2/(sqrt(3x^4+2)) to be 4/(sqrt(3+ g(x))? what is g(x)?

### 3 Answers

- PhilipLv 61 month agoFavorite Answer
Put 4x^2/[(3x^4+2)^(1/2)] = 4/{3+g}^(1/2)...(1), where g = g(x).;

Let's divide both numerator and denominator of LS(1) and get following::

LS(1) = 4/Q, where Q = (1/x^2)(3x^4+2)^(1/2) = [(3x^4+2)/(x^4)]^(1/2).;

Then Q = {3 + 2/x^4}^(1/2).;

For LS(1) = RS(1) we require 4/Q = 4/{3+g}^(1/2), ie., require {3+g}^(1/2) = Q.;

Then {3+g}^(1/2) = {3 +(2/x^4)}^(1/2), ie., g = 2/x^4.

- MorningfoxLv 71 month ago
We want x^2 / sqrt(3x^4 +2) to = 1/ sqrt( 3 + g(x))

So then sqrt(3x^4 + 2) / x^2 = sqrt( 3 + g(x))

which is the same as ...

sqrt( ( 3x^4 + 2) / x^4) = sqrt( 3 + g(x))

that tells us that ..

( 3x^4 + 2) / x^4 = 3 + g(x)

Subtract 3 from each side, to get

( 3x^4 + 2) / x^4 - 3 = g(x)

And that is g(x). Done!

- 1 month ago
4x^2 / sqrt(3x^4 + 2) = 4/sqrt(3 + g(x))

x^2 / sqrt(3x^4 + 2) = 1 / sqrt(3 + g(x))

sqrt(3x^4 + 2) / x^2 = sqrt(3 + g(x))

(3x^4 + 2) / x^4 = 3 + g(x)

3 + (2/x^4) = 3 + g(x)

2/x^4 = g(x)