Rationalize the following fraction:

$\displaystyle \frac{15}{\sqrt{5}}$

To rationalize a denominator, we will multiply the top and bottom of the fraction by the denominator.

$\displaystyle \frac{15}{\sqrt{5}}*\frac{\sqrt{5}}{\sqrt{5}}=\frac{15\sqrt{5}}{(\sqrt{5})^2}=\frac{15\sqrt{5}}{5}=3\sqrt{5}$

And we have our answer

First we need to factor both polynomials.

$\displaystyle \frac{x^{2}-9}{x^{2}-x-12}$

Becomes

$\displaystyle \frac{(x+3)(x-3)}{(x+3)(x-4)}$

Now we cancel out any variables that are in BOTH the numerator and denominator. Remember that if a group of variables/numbers are inside parenthesis, they are considered a single term.

The common term in this case is $\displaystyle (x+3)$, removing that from the equation gives us

$\displaystyle \frac{x-3}{x-4}$

The first step is to factor both polynomials:

$\displaystyle \frac{2b^{2}+7b+5}{b^{2}-1}$  Becomes $\displaystyle \frac{(2b+5)(b+1)}{(b+1)(b-1)}$

Now we cancel out any terms that are in both the numerator and denominator. Remember that any variables/numbers that are in parenthesis are considered a single term.

In this case, the common term is $\displaystyle (b+1)$

Once we remove that, we are left with:

$\displaystyle \frac{2b+5}{b-1}$

The first step is to factor both polynomials

$\displaystyle \frac{x^{2}-2x-8}{x^{2}-x-12}$   becomes $\displaystyle \frac{(x-4)(x+2)}{(x-4)(x+3)}$

Now we cancel out any terms that are in both the numerator and denominator. Remember that any variables/numbers that are in parenthesis are considered a single term.

In this case, the common term is $\displaystyle (x-4)$

Simplified, the equation is:

$\displaystyle \frac{x+2}{x+3}$

The first step is to factor both polynomials

$\displaystyle \frac{3x^{2}+11x+10}{3x^{2}-x-10}$ becomes $\displaystyle \frac{(3x+5)(x+2)}{(3x+5)(x-2)}$

Now we cancel out any terms that are in both the numerator and denominator. Remember that any variables/numbers that are in parenthesis are considered a single term.

In this case, the common term is $\displaystyle (3x+5)$

Simplified, the equation is:

$\displaystyle \frac{x+2}{x-2}$

The first step is to factor both polynomials

$\displaystyle \frac{16x^{2}-9}{4x^{2}-17x-15}$   becomes $\displaystyle \frac{(4x-3)(4x+3)}{(4x+3)(x-5)}$

Now we cancel out any terms that are in both the numerator and denominator. Remember that any variables/numbers that are in parenthesis are considered a single term.

In this case, the common term is $\displaystyle (4x+3)$

Simplified, the equation is:

$\displaystyle \frac{(4x-3)}{(x-5)}$

The first step is to factor both polynomials

$\displaystyle \frac{x^{2}-1}{x^{2}+3x-4}$ becomes $\displaystyle \frac{(x-1)(x+1)}{(x+4)(x-1)}$

Now we cancel out any terms that are in both the numerator and denominator. Remember that any variables/numbers that are in parenthesis are considered a single term.

In this case, the common term is $\displaystyle (x-1)$


Simplified, the equation is:

$\displaystyle \frac{(x+1)}{(x+4)}$

The first step is to factor both polynomials

$\displaystyle \frac{12x^{2}+11x+2}{6x^{2}+x-2}$ becomes  $\displaystyle \frac{(3x+2)(4x+1)}{(2x-1)(3x+2)}$

Now we cancel out any terms that are in both the numerator and denominator. Remember that any variables/numbers that are in parenthesis are considered a single term.

In this case, the common term is $\displaystyle (3x+2)$


Simplified, the equation is:

$\displaystyle \frac{(4x+1)}{(2x-1)}$

The first step is to factor both polynomials

$\displaystyle \frac{2x^{2}+x-15}{3x^{2}+4x-15}$ becomes  $\displaystyle \frac{(2x-5)(x+3)}{(3x-5)(x+3)}$

Now we cancel out any terms that are in both the numerator and denominator. Remember that any variables/numbers that are in parenthesis are considered a single term.

In this case, the common term is $\displaystyle (x+3)$


Simplified, the equation is:

$\displaystyle \frac{(2x-5)}{(3x-5)}$

The first step is to factor both polynomials

$\displaystyle \frac{5x^{2}+31x+6}{5x^{2}-29x-6}$  becomes    $\displaystyle \frac{(5x+1)(x+6)}{(5x+1)(x-6)}$

Now we cancel out any terms that are in both the numerator and denominator. Remember that any variables/numbers that are in parenthesis are considered a single term.

In this case, the common term is $\displaystyle (5x+1)$


Simplified, the equation is:

$\displaystyle \frac{(x+6)}{(x-6)}$