Multiplying and Dividing Rational Expressions: Additional Practice (Set 1)

Factor all four quadratics.

$x^2-5x+6 = (x-2)(x-3), \quad x^2-9 = (x-3)(x+3)$ $x^2+6x+9 = (x+3)^2, \quad x^2-4 = (x-2)(x+2)$

Multiply across, then cancel $x-3$, $x-2$, and one factor of $x+3$.

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

The restrictions are $x \ne 3$, $x \ne -3$, $x \ne 2$, and $x \ne -2$.

Factor each piece, pulling out the common $x$ where it appears.

$x^2-2x = x(x-2), \quad x^2+4x+4 = (x+2)^2, \quad x^2+2x = x(x+2), \quad x^2-4 = (x-2)(x+2)$

Multiply across and cancel $x-2$ and one factor of $x+2$.

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

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

Factor, then write the divisor as $\frac{x-5}{1}$ and flip it.

$x^2-7x+10 = (x-5)(x-2), \qquad x^2+2x = x(x+2)$

The denominator gives $x \ne 0$ and $x \ne -2$, and the divisor may not be zero, so $x \ne 5$.

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

The factor $x-5$ canceled, so the answer no longer mentions $x \ne 5$, yet the original expression is genuinely undefined there.

Rewrite $4-x^2$ so the factor $x-2$ appears, and factor the quadratic below it.

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

Multiply across. Every factor cancels, and the minus sign survives.

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

The answer is the constant $-1$, with $x \ne -3$, $x \ne -2$, and $x \ne 2$.

Factor both quadratics.

$x^2-2x-8 = (x-4)(x+2), \qquad x^2-9 = (x-3)(x+3)$

The dividend's denominator gives $x \ne 4$ and $x \ne -2$. The divisor's denominator gives $x \ne -4$. The divisor's numerator may not be zero, so $x \ne 3$ and $x \ne -3$.

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

The simplified form shows only $x \ne 4$, $x \ne -2$, and $x \ne -3$, losing both $x \ne -4$ and $x \ne 3$.

Factor $x^2+x-6 = (x+3)(x-2)$, then flip the divisor and cancel.

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

That formula is valid only where the original is defined: $x \ne 1$ from the two denominators, and $x \ne -3$ because the divisor may not be zero. Since $4$ is in the domain, $H(4) = 4 - 2 = 2$.

$H(1) \text{ requires } \frac{1^2+1-6}{1-1} = \frac{-4}{0},$

which is undefined, so $H(1)$ does not exist even though the formula $x-2$ would happily return $-1$.

Factor all four quadratics.

$x^2+3x-10 = (x+5)(x-2), \quad x^2+x-6 = (x+3)(x-2)$ $x^2-25 = (x-5)(x+5), \quad x^2-9 = (x-3)(x+3)$

Flip the divisor and multiply, canceling $x+5$, $x-2$, and $x+3$.

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

The restrictions are $x \ne -5$, $x \ne -3$, $x \ne 2$, $x \ne 3$, and $x \ne 5$.

The algebra is right: factoring and flipping gives

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

But the restrictions belong to the original expression, not to the answer. Both denominators are $x+2$, so $x \ne -2$, and the divisor $\frac{x-3}{x+2}$ is zero at $x = 3$, where dividing by it is undefined, so $x \ne 3$.

The polynomial $x+3$ is defined everywhere and therefore records neither fact. The complete answer is $x+3$ with $x \ne -2$ and $x \ne 3$.

Factor all four polynomials.

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

Multiply across, then cancel one factor of $x-2$ and one factor of $x+2$.

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

The restrictions are $x \ne 2$, $x \ne 0$, and $x \ne -2$.

Factor the cubic by grouping.

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

With $x^2+x-6 = (x+3)(x-2)$ underneath, multiply across and cancel $x+3$, $x-2$, and $x+2$.

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

The restrictions are $x \ne -3$, $x \ne 2$, and $x \ne -2$, none of which the answer displays.

Take the three conditions one at a time. The dividend's denominator gives $x^2 - 4 \ne 0$, so $x \ne 2$ and $x \ne -2$. The divisor's denominator gives $x^2 + 4 \ne 0$, which costs nothing, since $x^2 + 4 \ge 4$ for every real $x$. The divisor's numerator may not be zero.

$x^2 - 9 = 0 \;\Rightarrow\; x = 3 \text{ or } x = -3$

That gives the excluded set $\{-3, -2, 2, 3\}$, which has $4$ elements.

The first numerator is a perfect square.

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

Multiply across, then cancel $x+1$ and one factor of $x-1$.

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

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