Algebra II - Multiplying and Dividing Rational Expressions

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

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

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

$\frac{x}{x+1}$

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

Explanation

Rewrite the left fraction using common factors.

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

Cancel out common terms.

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

Factorize the bottom term.

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

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

Simplify: $\frac{x}{x-9}\div\frac{x-9}{x-18}$

$\frac{x^2-18}{x^2-18+81}$

$\frac{x+9}{x^2-18+81}$

$\frac{x}{x-18}$

$\frac{x-18}{x}$

$\frac{x+2}{x-9}$

Explanation

In order to simplify the rational expression, we will need to rewrite the expression as a multiplication sign and take the reciprocal of the second term.

$\frac{x}{x-9}\div\frac{x-9}{x-18}= \frac{x}{x-9} \times \frac{x-18}{x-9}$

Simplify the numerator.

Simplify the denominator by FOIL method.

Divide the numerator with the denominator.

The answer is: $\frac{x^2-18}{x^2-18+81}$

(9_x_2 – 1) / (3_x_ – 1) =

3_x_ + 1

3_x_ – 1

(3_x_ – 1)2

3_x_

Explanation

It's much easier to use factoring and canceling than it is to use long division for this problem. 9_x_2 – 1 is a difference of squares. The difference of squares formula is a_2 – b_2 = (a + b)(a – b). So 9_x_2 – 1 = (3_x + 1)(3_x – 1). Putting the numerator and denominator together, (9_x_2 – 1) / (3_x_ – 1) = (3_x_ + 1)(3_x_ – 1) / (3_x_ – 1) = 3_x_ + 1.

Which is a simplified form of $\frac{2a^{2}b^{3}c^{-2}}{(4a^{-1}b^{2}c)^{3}}$ ?

$\frac{a^5b^{-3}c^{-5}}{32}$

$\frac{a^{-1}b^{-3}c^{-5}}{32}$

$\frac{a^3b^{-3}c^{-5}}{32}$

$\frac{a^5b^{-3}c}{32}$

$\frac{a^{-1}b^{-3}c^{5}}{32}$

Explanation

$\frac{2a^{2}b^{3}c^{-2}}{(4a^{-1}b^{2}c)^{3}}$

$=\frac{2a^{2}b^{3}c^{-2}}{4^{3}a^{-1\times 3}b^{2\times 3}c^{3}}$

$=\frac{2a^{2}b^{3}c^{-2}}{64a^{-3}b^{6}c^{3}}$

$=\frac{1}{32}\times a^{2-(-3)}\times b^{3-6}\times c^{-2-3}$

$=\frac{a^5b^{-3}c^{-5}}{32}$

Multiply:

$\frac{2s^3(9s^2-25)}{s+1}*\frac{3s^2+8s+5}{12s^2-20s}$

$\frac{9}{2}s^4+15s^3+\frac{25}{2}s^2$

$\frac{1}{2}s^2$

$\frac{1}{2}s^2+\frac{9}{2}x^2-\frac{25}{2}$

Explanation

First, completely factor everything that can possibly be factored. This includes both numerators and the second denominator:

$\frac{2s*s*s(3s+5)(3s-5)}{s+1}*\frac{(s+1)(3s+5)}{2*2s(3s-5)}$

Now we can cancel everything that appears both on the top and the bottom, since it will divide to be a factor of :

$\frac{s*s(3s+5)}{1}*\frac{(3s+5)}{2}$

We can simplify this by multiplying and .

$\frac{s*s(3s+5)(3s+5)}{1*2}=\frac{s^{2}(9s^2 + 15s + 25)}{2}$

This leaves us with the following answer:

$\frac{9}{2}s^4+15s^3+\frac{25}{2}s^2$

$\frac{\frac{4}{x}}{\frac{6}{x^2}}$

$\frac{2x}{3}$

$\frac{2}{3}$

$\frac{x}{3}$

$\frac{1}{3}$

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

Explanation

In this problem, we're dealing with dividing rational expressions. Therefore, we have to flip the second fraction and then multiply the two: $\frac{4}{x}\cdot \frac{x^2}{6}$ . Simplify and multiply straight across to get your answer: $\frac{2x}{3}$ .

$\frac{2a}{b^2}\cdot \frac{5b^4}{10a}$

$\frac{1}{b^2}$

Explanation

When multiplying fractions, you will multiply straight across.

But first, see if you can reduce diagonally.

The a's cross out, and you can take out a from the other diagonal.

The coefficients also reduce.

$\frac{2a}{b^2}\cdot \frac{5b^4}{10a}=\frac{2}{1}\cdot \frac{b^2}{2}=b^2$

Therefore, your answer is .

$\frac{2m^2+13m+15}{2m^2-5m-12}*\frac{3m^2-14m+8}{3m^2+13m-10}$

$\frac{m+5}{m-4}$

$\frac{1}{m+5}$

Explanation

First, completely factor all 4 quadratics:

$\frac{(m+5)(2m+3)}{(2m+3)(m-4)}*\frac{(m-4)(3m-2)}{(m+5)(3m-2)}$

Now we can cancel all factors that appear on both the top and the bottom, because those will divide to a factor of . We quickly realize that all of the factors can be crossed off. This means that all of the factors have been divided to . This leves us with the following answer: