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# Manipulating Rational Expressions

## Manipulating Rational Expressions 1.1 - Solution

a
We can simplify the expression if we can factor out the same factors in the numerator and the denominator, and then let them cancel each other out. All terms in the numerator and also the denominator is divisible by $4$ so we can factor this factor out.
$\dfrac{16+8x}{4x-4x^2}$
$\dfrac{4\cdot 4+4\cdot 2x}{4\cdot x-4\cdot x^2}$
$\dfrac{4\left(4+2x\right)}{4\left(x-x^2\right)}$
$\dfrac{4+2x}{x-x^2}$
Even if we can still factor out a $2$ in the numerator and $x$ in the denominator, these factors can not cancel each other out.
b
All terms contain $x.$ This means that we can factor out the $x$ in both the numerator and the denominator and have them cancel each other out.
$\dfrac{x^2-2x}{x^3+3x}$
$\dfrac{x\cdot x-x\cdot 2}{x\cdot x^2+x\cdot 3}$
$\dfrac{x\left(x-2\right)}{x\left(x^2+3\right)}$
$\dfrac{x-2}{x^2+3}$
c
It is not clear which factors of the numerator and denominator have in common. Let's start by factoring out $3.$
$\dfrac{3x-9}{x^2-3x}$
$\dfrac{3\cdot x-3\cdot 3}{x\cdot x-x\cdot 3}$
$\dfrac{3(x-3)}{x(x-3)}$
Now we see that the numerator and the denominator share a factor $x-3.$
$\dfrac{3(x-3)}{x(x-3)}$
$\dfrac{3}{x}$