College Algebra - Rational Expressions

Simplify the following:

$\frac{12x^{2}+11x+2}{6x^{2}+x-2}$

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

$\frac{(4x+1)}{(3x-1)}$

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

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

$\frac{(4x+1)}{(3x+2)}$

Explanation

The first step is to factor both polynomials

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

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

Simplified, the equation is:

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

Simplify the following:

$\frac{5x^{2}+31x+6}{5x^{2}-29x-6}$

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

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

$\frac{(x+6)}{(5x-6)}$

$\frac{(5x+6)}{(x-6)}$

$\frac{(5x+6)}{(3x-6)}$

Explanation

The first step is to factor both polynomials

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

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

Simplified, the equation is:

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

Simplify the following:

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

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

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

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

$\frac{x+3}{x+4}$

$\frac{x-2}{x-4}$

Explanation

First we need to factor both polynomials.

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

Becomes

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

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 , removing that from the equation gives us

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

Simplify the following

$\frac{2b^{2}+7b+5}{b^{2}-1}$

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

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

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

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

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

Explanation

The first step is to factor both polynomials:

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

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

Once we remove that, we are left with:

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

Simplify the following:

$\frac{16x^{2}-1}{4x^{2}-15x-4}$

$\frac{(4x-1)}{(x-4)}$

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

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

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

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

Explanation

The first step is to factor both polynomials

$\frac{16x^{2}-1}{4x^{2}-15x-4}$ becomes $\frac{(4x+1)(4x-1)}{(4x+1)(x-4)}$

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

Simplified, the equation is:

$\frac{(4x-1)}{(x-4)}$

Add: $\frac{1}{x}+\frac{2}{x-1}$

$\frac{3x-1}{x^2-x}$

$\frac{3x-1}{x^2-1}$

$\frac{3}{x^2-x}$

$-\frac{x-1}{x^2-x}$

$-\frac{x-3}{x^2-x}$

Explanation

To add the terms of the numerator, we will need to have the same denominator.

Multiply both denominators together.

Multiply the numerators with what was multiplied on the denominator.

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

Combine like terms.

The answer is: $\frac{3x-1}{x^2-x}$

Simplify the following:

$\frac{x^{2}-2x-8}{x^{2}-x-12}$

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

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

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

$\frac{x-2}{x-3}$

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

Explanation

The first step is to factor both polynomials

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

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

Simplified, the equation is:

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

Simplify the following:

$\frac{2x^{2}+x-15}{3x^{2}+4x-15}$

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

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

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

$\frac{(3x+5)}{(3x-5)}$

$\frac{(2x-5)}{(2x+5)}$

Explanation

The first step is to factor both polynomials

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

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

Simplified, the equation is:

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

Simplify the following:

$\frac{6x^{2}-x-1}{12x^{2}+x-1}$

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

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

$\frac{(3x+1)}{(4x-1)}$

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

$\frac{(3x+1)}{(3x-1)}$

Explanation

The first step is to factor both polynomials

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

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