Simplifying $\frac{3(a^{1/4}+4)}{2a-32a^{1/2}}$

Question

I have a fraction $\frac{3a^{1/4}+12}{2a-32a^{1/2}}$ which I have factored out into $\frac{3\left(a^{\frac{1}{4}}+4\right)}{2a-32a^{\frac{1}{2}}}$, but checking out W|A I also get that there ought to be also $\frac{3}{2(a^{1/4}-4)a^{1/2}}$

I can't figure out what steps should I take between what I've got and that simpler form though.

Answer 1

$$ 2a-32a^{1/2} = 2\left(a -16a^{1/2}\right) = 2a^{1/2}\left(a^{1/2}-16\right) $$ Since this is of the form $a^2-b^2$ we can factor as follows $$ 2a^{1/2}\left(a^{1/2}-16\right) = 2a^{1/2}\left(a^{1/4}-4\right)\left(a^{1/4}+4\right) $$ I think that is enough to get through this.

Answer 2

Note that $$\begin{align}2a-32a^{\frac 12}&=2a^{\frac 12}(a^{\frac 12}-16)\\&=2a^{\frac 12}\left(\left(a^{\frac 14}\right)^2-4^2\right)\\&=2a^{\frac 12}\left(a^{\frac 14}-4\right)\left(a^{\frac 14}+4\right).\end{align}$$

Answer 3

In the denominator of your fraction: $$2a-32a^{\frac 12 }=2a^{\frac 12}\left(a^{\frac{1}{2}}-16\right)=2a^{\frac 12}\left(\left(a^{\frac 14}\right)^2-4^2\right)$$

and use $x^2-y^2=(x-y)(x+y)$

Answer 4

i will make a change of variable $x = a^{1/4}, a = x^4.$ with this, we have $$\frac{3(x+4)}{2x^4 - 32x^2} = \frac{3(x+4)}{2x^2(x^2 - 16)}=\frac 3{2x^2(x-4)} = \frac 3{2a^{1/2}(a^{1/4}-4)}= \frac 3{2a^{1/2}(a^{1/16}-\sqrt 2)(a^{1/16}+\sqrt 2)(a^{1/8}+2)}$$