# How do you simplify (x^2 – 3x – 10)/(x^2 – 9)div(x^2 + 4x + 4)/(x+3)?

May 21, 2015

First, factor your quadratic functions. Finding its roots will lead us to the factors and then it'll be way easier to calculate your result. So, let's find the roots (using Bhaskara)!

$f \left(x\right) = {x}^{2} - 3 x - 10$

$\frac{3 \pm \sqrt{9 - 4 \left(1\right) \left(- 10\right)}}{2}$
$\frac{3 \pm 7}{2}$
${x}_{1} = 5$, the same as the factor $x - 5 = 0$
${x}_{2} = - 2$, the same as the factor $x + 2 = 0$

$f \left(x\right) = {x}^{2} - 3 x - 10 = \left(x - 5\right) \left(x + 2\right)$

Next: $g \left(x\right) = {x}^{2} - 9$

${x}^{2} = 9$
$x = \pm 3$, which generate the factors $\left(x - 3\right) \left(x + 3\right)$

Finally, $h \left(x\right) = {x}^{2} + 4 x + 4$, which will have one root: $x = - 2$ and then give us the same factor twice: $\left(x + 2\right) \left(x + 2\right)$

Rewriting:

$\frac{\frac{\left(x - 5\right) \left(x + 2\right)}{\left(x - 3\right) \left(x + 3\right)}}{\frac{\left(x + 2\right) \left(x + 2\right)}{x + 3}}$=$\frac{\left(x - 5\right) \cancel{\left(x + 2\right)} \cancel{\left(x + 3\right)}}{\left(x - 3\right) \cancel{\left(x + 3\right)} \cancel{\left(x + 2\right)} \left(x + 2\right)}$=$\frac{x - 5}{\left(x - 3\right) \left(x + 2\right)}$=$\frac{x - 5}{{x}^{2} - x - 6}$

May 21, 2015

This explanation's going to be all math. I'm assuming you know how to factor polynomials, but if you don't, then don't be afraid to comment to ask how; I don't bite! Okay, let's start.

(x^2–3x–10)/(x^2–9) ÷ (x^2+4x+4)/(x+3)

We can flip the 2nd fraction and multiply it.

(x^2–3x–10)/(x^2–9) times (x+3)/(x^2+4x+4)#

Now let's factor the trinomials into binomials. Hopefully this will let us cancel out some stuff.

$\frac{\left(x - 5\right) \left(x + 2\right)}{{x}^{2} - 9} \times \frac{x + 3}{\left(x + 2\right) \left(x + 2\right)}$

We can also factor ${x}^{2} - 9$ into $x + 3$ and $x - 3$.

$\frac{\left(x - 5\right) \left(x + 2\right)}{\left(x - 3\right) \left(x + 3\right)} \times \frac{x + 3}{\left(x + 2\right) \left(x + 2\right)}$

Now, I can cancel out some of the factors to simplify the answer.

$\frac{x - 5}{x - 3} \times \frac{1}{x + 2} = \frac{x - 5}{\left(x - 3\right) \left(x + 2\right)} = \frac{x - 5}{{x}^{2} - x - 6}$

This should be the final answer, unless your teacher prefers to show the factors of trinomials, in which case $\frac{x - 5}{\left(x - 3\right) \left(x + 2\right)}$ is also acceptable.