# How to factor this rational expression?

$$x^2 + {16\over x^2} -12$$

This is what I've done so far... $$x^4 -12x^2 +16$$

$$= (x^2 - 4)(x^2 + 4) -12x^2$$

What shall I do next?

EDIT: Thanks for the answer, however the answer given here is this: $$\left({x^2+2x-4 \over x}\right)\left({x^2-2x-4 \over x}\right)$$ I'd like to know how the above is achieved from the question.

-
It's not clear to me what you're looking for. Usually factoring is something you do to polynomials (not necessarily, though) and the expression $x^2+16x^{-2}-12$ can never be a polynomial on $x$. – Git Gud Apr 21 '13 at 10:04
@GitGud Surely you can factor rational functions too. – Abel Apr 21 '13 at 10:07

Given expression, $$x^{2}+\left(\frac{4}{x}\right)^{2}-2\cdot\frac{4}{x}\cdot x-4$$

$$=\left(x-\frac{4}{x}\right)^{2}-2^{2}$$

$$=\left(x-\frac{4}{x}-2\right)\left(x-\frac{4}{x}+2\right)$$

$$=\left({x^2+2x-4 \over x}\right)\left({x^2-2x-4 \over x}\right)$$

-
How do you get $2 * {4x \over x} -4$? – Ghost Apr 21 '13 at 10:22
$-12 = -8-4$ and $2*\frac{4x}{x}=8$.So $-12=-2*\frac{4x}{x}-4$ – Shaswata Apr 21 '13 at 10:26

Step-by-step: "How'd you get that?" \begin{align} x^2 + \dfrac{16}{x^2} - 12 & \iff (x)^2 + \left(\frac{4}{x}\right)^2 - \color{green}{\bf 8}- 4 \\ \\ & \iff (x)^2 - \color{green}{\bf 8 \cdot\dfrac xx} + \left(\frac 4x\right)^2 - 4 \\ \\ & \iff (x)^2 - \color{green}{\bf 2\cdot\dfrac{4x}{x}} + \left(\frac 4x\right)^2 - 4 \\ \\ & \iff (\color{red}{\bf x})^2 - 2\cdot \left(\frac{\color{blue}{\bf 4}\color{red}{\bf x}}{\color{blue}{\bf x}}\right) + \left(\color{blue}{\bf \frac{4}{x}}\right)^2 - 4 \\ \\ & \iff \left(x - \dfrac{4}{x}\right)^2 - 4 \qquad\qquad\qquad\qquad\qquad \tag{\color{red}{\bf a}^2 - 2\color{red}{\bf a}\color{blue}{\bf b} + \color{blue}{\bf b}^2 = (\color{red}{\bf a} - \color{blue}{\bf b})^2} \\ \\ & \iff \left(x-\frac{4}{x}\right)^2 - 2^2 \tag{difference of squares}\\ \\ & \iff \left(\left(x - \frac{4}{x}\right) + 2\right)\left(\left(x - \dfrac{4}{x}\right) - 2 \right)\\ \\ & \iff \left(x -\frac{4}{x}+ 2\right)\left(x - \frac 4x - 2\right)\tag{factored} \\ \\ & \iff \left(\frac{x^2 - 4 + 2x}{x}\right)\left(\frac{x^2 - 4 - 2x}{x}\right) \qquad\qquad\qquad\tag{common denominator} \end{align}

-
I always like seeing clear answers and color coding is so nice! +1 – Amzoti Apr 21 '13 at 17:13
I hope this helps clarify the factorization, Ghost! ;-) – amWhy Apr 22 '13 at 2:30

When you have $x^4-12x^2+16$ you can directly solve it and get a factorization.

If you want to get it without solving, you can form a difference of two squares : $x^4 - 12x^2 + 16 = (x^2-6)^2 - (2 \sqrt{5})^2 = (x^2 - 6 - 2\sqrt{5})(x^2 - 6 + 2 \sqrt{5})$ then you only need to find the square roots of $6 \pm 2\sqrt{5}$.

-

Factoring means, in the simple case with three factors, that we have the equality $$x^2+16/x^2-12 = (x^2+a)(x^2+b)/x^2$$ with some unknown a and b.

Reordering gives $$x^4+16-12x^2 = (x^2+a)(x^2+b)$$ where the OP has already found the lhs. Now we search for a and b. $$x^4-12x^2+16 = x^4+(a+b)x^2 +ab \\ -12x^2+16 = (a+b)x^2 +ab$$ To have this correct for every admissible value of x, the coefficients at the powers of x must be equal; thus we get now the two equations to solve simultaneously: $$-12 = a+b \\ 16 = ab$$ The first one says $b=-(a+12)$ and we insert this into the second one, giving $$16 = -a(a+12) = -a^2-12a$$ or $$-16 = a^2+12a$$ Here we introduce now the well known technique of the quadratic supplement $(12/2)^2 = 6^2$ $$-16+6^2 = a(a-12) = a^2+12a +6^2$$ giving $$20 = (a+6)^2$$ from which follows $$a= -6+\sqrt{20}$$ and then $$b= -(( -6+\sqrt{20})+12) = -6-\sqrt{20}$$

-