Rational Expressions - College Algebra

Card 0 of 17

Question

Rationalize the following fraction:

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

Answer

Rationalize the following fraction:

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

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

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

And we have our answer

Compare your answer with the correct one above

Question

Simplify the given expression:

$\frac{x^2+4x-12}{2x+12}$

Answer

First we want to factor both the numerator and denominator:

$\frac{x^2+4x-12}{2x+12}$

For the numerator, we can use the quadratic formula:

$x=\frac{-(4)\pm \sqrt{4^{2}-4(1)(-12)}}{2(1)}$

$\x=\frac{-4\pm \sqrt{16+48}}{2}\x=\frac{-4\pm\sqrt{64}}{2}\x=\frac{-4\pm8}{2}$

$\x=\frac{-12}{2};or;x=\frac{4}{2}\\x=-6;or;x=2$

For the denominator, we only need to factor out the , leaving us with the given factored expression:

$\frac{(x+6)(x-2)}{2(x+6)}$

Then we cancel out like terms, giving us a simplified expression:

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

Compare your answer with the correct one above

Question

Simplify the following:

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

Answer

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

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

Compare your answer with the correct one above

Question

Simplify the following

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

Answer

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:

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

Compare your answer with the correct one above

Question

Simplify the following:

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

Answer

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:

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

Compare your answer with the correct one above

Question

Simplify the following:

$\frac{3x^{2}+11x+10}{3x^{2}-x-10}$

Answer

The first step is to factor both polynomials

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

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

Simplified, the equation is:

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

Compare your answer with the correct one above

Question

Simplify the following:

$\frac{16x^{2}-9}{4x^{2}-17x-15}$

Answer

The first step is to factor both polynomials

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

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

Simplified, the equation is:

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

Compare your answer with the correct one above

Question

Simplify the following:

$\frac{2x^{2}+13x-7}{2x^{2}-17x+8}$

Answer

The first step is to factor both polynomials

$\frac{2x^{2}+13x-7}{2x^{2}-17x+8}$ becomes $\frac{(2x-1)(x+7)}{(2x-1)(x-8)}$

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:

$\frac{(x+7)}{(x-8)}$

Compare your answer with the correct one above

Question

Simplify the following:

$\frac{x^{2}-1}{x^{2}+3x-4}$

Answer

The first step is to factor both polynomials

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

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

Simplified, the equation is:

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

Compare your answer with the correct one above

Question

Simplify the following:

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

Answer

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:

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

Compare your answer with the correct one above

Question

Simplify the following:

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

Answer

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:

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

Compare your answer with the correct one above

Question

Simplify the following:

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

Answer

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:

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

Compare your answer with the correct one above

Question

Simplify the following:

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

Answer

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

Simplified, the equation is:

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

Compare your answer with the correct one above

Question

Simplify the following:

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

Answer

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:

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

Compare your answer with the correct one above

Question

Simplify: $\frac{3x^5}{2x^5+x}$

Answer

Factor an in the denominator.

$\frac{3x^5}{2x^5+x} = \frac{3x^5}{x(2x^4+1)}$

Cancel the common in the numerator and denominator.

The answer is: $\frac{3x^4}{2x^4+1}$

Compare your answer with the correct one above

Question

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

Answer

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

Compare your answer with the correct one above

Question

Add/Subtract: $\frac {4}{5}-\frac {1}{4}$ $\frac {4}{5}-\frac {1}{4}+1$

Answer

Step 1: Find the Least Common denominator of all the fractions..

LCD=

Step 2: Convert all fractions to denominator .

$\frac {4}{5}\equiv \frac {16}{20}$

$\frac {1}{4}\equiv \frac {5}{20}$

$1=\frac {20}{20}$

Step 3: Re-write all fractions in terms of the original equation...

$\frac {16}{20}-\frac {5}{20}+\frac {20}{20}$

Answer: $\frac {31}{20}$ or $1\frac {11}{20}$

Compare your answer with the correct one above

Tap the card to reveal the answer