1. ## Simplifying Complex Fractions

Hi, I have a fraction which I can't simplify fully.

I know that becomes (x + 2)(x - 2)

By solving with a calculator, the answer is:

How do I get there, with working?

Thanks, BG

2. Originally Posted by BG5965
Hi, I have a fraction which I can't simplify fully.

I know that becomes (x + 2)(x - 2)

By solving with a calculator, the answer is:

How do I get there, with working?

Thanks, BG

the numerator can be simplified to $x^2+1$

$2x^2+x+4+\frac{1}{x}+\frac{2}{x^2}=0$

$2x^2+\frac{2}{x^2}+x+\frac{1}{x}+4=0$

$2(x^2+\frac{1}{x^2})+x+\frac{1}{x}+4=0$

Let $y=x+\frac{1}{x}$

$
2(y-2)+y+4=0
$

$3y=0$, $y=0$thus $x+\frac{1}{x}=0$which implies $x^2+1=0$

Anyways , i am not sure about the denominator . Sorry about that ..

3. You can make the simplifying assumption that they've given you something "find-able" in the way of factors. (I'm sure you've checked already, and determined that the numerator has no real roots, and that the denominator has only irrational roots.)

Since the denominator has no linear term, then the constant terms and the linear terms in the factors must create terms that cancel out. So let's guess that the leading coefficients are "2", and that the constant terms are -2 and +2. Then let's see if we can find a product that works:

(2x^2 + ax + 2)(2x^2 + bx - 2)

= 4x^4 + (2a + 2b)x^3 + (ab)x^2 + (2b - 2a)x - 4

= 4x^4 + 4x^3 + x^2 - 4

Then 2b - 2a = 0, so 2b = 2a and thus a = b. This gives us:

4a = 4

...so a = 1. Checking:

a^2 = 1

...which matches the given coefficient of the squared term. Then the factors turn out to be:

(2x^2 + x + 2)
(2x^2 + x - 2)

Divide each of these into the numerator, and see which one comes out "even".

No, I don't see a way to do this in a more straightforward manner...