Multiplying and Dividing Rational Expressions: Practice

Factor, and read the restrictions off the original denominators while they are factored.

$x^2-16 = (x-4)(x+4), \qquad x^2-x-6 = (x-3)(x+2)$

The denominators are $(x-3)(x+2)$ and $x-4$, so $x \ne 3$, $x \ne -2$, and $x \ne 4$. Now multiply and cancel.

$\frac{(x-4)(x+4)}{(x-3)(x+2)} \cdot \frac{x+2}{x-4} = \frac{x+4}{x-3}$

The simplified form shows only $x \ne 3$. The other two restrictions rode on the factors $x+2$ and $x-4$, which cancelled, so they must be carried by hand.

Collect the restrictions from the original quotient first. Both denominators are $x+1$, so $x \ne -1$. The divisor $\frac{x-5}{x+1}$ is zero exactly when $x - 5 = 0$, and you cannot divide by zero, so $x \ne 5$.

Now flip and multiply, using $x^2-25 = (x-5)(x+5)$.

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

The answer is the polynomial $x+5$, which is defined everywhere and therefore advertises none of its restrictions. Both $x \ne -1$ and $x \ne 5$ are still true of the original expression.

Factor the numerator, and write the polynomial divisor over $1$.

$x^2-x-12 = (x-4)(x+3)$

Flip the divisor and multiply.

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

The restrictions are $x \ne -1$ from the denominator and $x \ne 4$ from the divisor, which may not be zero.

The first numerator is a perfect square and the first denominator is a difference of squares.

$x^2+6x+9 = (x+3)^2, \qquad x^2-9 = (x-3)(x+3)$

Multiply across, then cancel one $x+3$ and the $x-3$.

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

The restrictions are $x \ne 3$, $x \ne -3$, and $x \ne 0$.

Factor, and notice that $5-x$ is the opposite of $x-5$.

$x^2+2x = x(x+2), \qquad 5-x = -(x-5)$

Now multiply across. The factors $x-5$ and $x+2$ both cancel, and the minus sign survives.

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

The restrictions are $x \ne -2$ and $x \ne 5$.

A quotient carries three conditions. The dividend's denominator gives $x - 1 \ne 0$, so $x \ne 1$. The divisor's denominator gives $x - 6 \ne 0$, so $x \ne 6$. The divisor's numerator may not be zero either, since dividing by zero is undefined.

$x + 3 = 0 \;\Rightarrow\; x = -3 \quad \text{is excluded}$

The dividend's numerator $x+7$ is the one polynomial that is free to vanish, so $x = -7$ is allowed.

Factor both quadratics.

$2x^2+5x-3 = (2x-1)(x+3), \qquad x^2-9 = (x-3)(x+3)$

Multiply across and cancel $2x-1$ along with one copy of $x+3$.

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

The restrictions are $x \ne 3$, $x \ne -3$, and $x \ne \tfrac{1}{2}$.

The rule extends to any number of factors: multiply all the numerators and all the denominators.

$\frac{x(x-1)(x+2)}{(x-1)(x+2)(x+5)}$

The factors $x-1$ and $x+2$ each appear on both the top and the bottom, so both cancel.

$\frac{x(x-1)(x+2)}{(x-1)(x+2)(x+5)} = \frac{x}{x+5}$

The restrictions are $x \ne 1$, $x \ne -2$, and $x \ne -5$.

The simplified form $x-1$ would give $3 - 1 = 2$, but the question asks about the original expression, and at $x = 3$ its first denominator is zero.

$\frac{3^2-1}{3-3} = \frac{8}{0}$

Division by zero is undefined, so the original expression has no value at $x = 3$. Cancelling the factor $x-3$ removed the evidence, not the restriction: the answer is $x-1$ with $x \ne 3$ and $x \ne -1$.

Factor the numerator completely, pulling out $x$ first and then using the difference of squares.

$x^3 - x = x(x^2-1) = x(x-1)(x+1)$

Multiply across and cancel the factor $x$.

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

The restrictions are $x \ne -4$ and $x \ne 0$.

Factor the numerator, then write the divisor as $\frac{3x}{1}$ and flip it.

$6x^2+9x = 3x(2x+3)$

Multiplying by the reciprocal of $3x$ gives

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

The restrictions are $x \ne 2$ from the denominator and $x \ne 0$, because the divisor $3x$ may not be zero.

The original has denominators $x-7$ and $x+1$, so the restrictions are $x \ne 7$ and $x \ne -1$.

$\frac{x+2}{x-7} \cdot \frac{x-7}{x+1} = \frac{x+2}{x+1}$

The simplified form has only the denominator $x+1$, so it still shows $x \ne -1$ but has lost every trace of $x \ne 7$. That restriction rode on the factor $x-7$, which cancelled. ($x = -2$ merely makes a numerator zero, which is allowed.)