# How do you simplify (x^3 - 2x^2 - 3x)/(2x^2 - 6x)?

May 27, 2015

You can start by factoring out what's multiplying all the terms both in the numerator and denominator; then, we can find the roots of what's left and rewritethe fraction as a quotient of factors.

$\frac{\cancel{x} \left({x}^{2} - 2 x - 3\right)}{\cancel{x} \left(2 x - 6\right)}$

Now, let's find the roots of the quadratic:

$\frac{2 \pm \sqrt{4 - 4 \left(1\right) \left(- 3\right)}}{2}$
$\frac{2 \pm 4}{2}$

${x}_{1} = 3$, which, equaled to zero, turns into the factor $\left(x - 3\right) = 0$
${x}_{2} - 1$, which, equaled to zero, turns into the factor $\left(x + 1\right) = 0$

And about the denominator, we have that $2$ is multiplying both terms, so $\left(2 x - 6\right) = 2 \left(x - 3\right)$

Now, rewriting it all over again:

$\frac{\cancel{x - 3} \left(x + 1\right)}{2 \cancel{x - 3}}$

Thus, the final answer is $\frac{x + 1}{2}$