Multiplying and Dividing Rational Expressions: Practice

A product carries no hidden condition: the restrictions are exactly the ones its denominators impose. Set each denominator to zero.

$x - 3 = 0 \;\Rightarrow\; x = 3, \qquad x + 2 = 0 \;\Rightarrow\; x = -2$

So $x \ne 3$ and $x \ne -2$. The numerator $x$ is allowed to be zero, so $x = 0$ is not excluded.

Multiply across, keeping everything factored.

$\frac{x+4}{x-1} \cdot \frac{x-1}{x+7} = \frac{(x+4)(x-1)}{(x-1)(x+7)}$

The factor $x-1$ appears on the top and on the bottom, so it cancels.

$\frac{(x+4)(x-1)}{(x-1)(x+7)} = \frac{x+4}{x+7}$

The restrictions $x \ne 1$ and $x \ne -7$ are carried along from the original denominators.

The reciprocal of a fraction is the fraction turned upside down: the numerator and the denominator swap places.

$\text{reciprocal of } \frac{x-6}{x+2} \;=\; \frac{x+2}{x-6}$

A check confirms it, since the two multiply to $1$ whenever both are defined: $\frac{x-6}{x+2} \cdot \frac{x+2}{x-6} = 1$.

To divide, invert the divisor and multiply. The dividend $\frac{x+1}{x-4}$ stays exactly where it is; only the divisor $\frac{x+3}{x-5}$ is flipped.

$\frac{x+1}{x-4} \div \frac{x+3}{x-5} = \frac{x+1}{x-4} \cdot \frac{x-5}{x+3}$

Flipping the dividend, flipping both, or flipping neither all change the value of the expression.

The second factor is the reciprocal of the first, so every factor on the top also appears on the bottom.

$\frac{x-7}{x+5} \cdot \frac{x+5}{x-7} = \frac{(x-7)(x+5)}{(x+5)(x-7)} = 1$

The answer is $1$, valid for $x \ne -5$ and $x \ne 7$.

Only whole factors cancel. Here $x-3$ is a factor of both the numerator and the denominator, so it cancels.

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

The $x$ inside $x+2$ and the $x$ inside $x+5$ are terms, not factors, so they cannot be cancelled against each other; that mistake would give the false answer $\frac{2}{5}$.

Factor the difference of squares first.

$x^2 - 9 = (x-3)(x+3)$

Now multiply across and cancel the factors $x+3$ and $x-3$.

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

The restrictions $x \ne 3$ and $x \ne -3$ are carried along.

Read the restrictions off the original denominators, before any cancelling. The first denominator $x(x-4)$ vanishes at $x = 0$ and $x = 4$; the second vanishes at $x = -6$.

$\frac{x+1}{x(x-4)} \cdot \frac{x-4}{x+6} = \frac{x+1}{x(x+6)}$

The factor $x-4$ cancels, so the simplified form no longer shows $x \ne 4$. The restriction is still real: the original expression divides by $x-4$.

Flip the divisor and multiply.

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

Anything nonzero divided by itself is $1$, and the divisor is nonzero exactly when $x \ne -2$. Together with $x \ne 1$ from the denominators, the answer is $1$ for $x \ne -2$ and $x \ne 1$.

Rational expressions multiply exactly like fractions, so the tops multiply and the bottoms multiply.

$\frac{A}{B} \cdot \frac{C}{D} = \frac{AC}{BD}$

A common denominator is a tool for adding and subtracting, not multiplying, and inverting the second expression is what you do to divide, not to multiply.

Write the polynomial as a rational expression with denominator $1$, then multiply across.

$\frac{2}{x-8} \cdot \frac{x-8}{1} = \frac{2(x-8)}{x-8}$

The factor $x-8$ cancels.

$\frac{2(x-8)}{x-8} = 2$

The answer is the constant $2$, with $x \ne 8$ carried along from the original denominator.

The denominator $x-1$ gives the first restriction.

$x - 1 = 0 \;\Rightarrow\; x = 1$

The divisor is the polynomial $x+4$, and you may never divide by zero, so $x + 4 \ne 0$, which gives $x \ne -4$. Nothing in the expression as written has $x+4$ in a denominator, so this second restriction comes from the division symbol alone.