Multiplying and Dividing Rational Expressions: Practice

Factor, and rewrite $9-x^2$ so that the factor $x-3$ appears.

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

Flip the divisor and multiply.

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

The answer is the constant $-1$, valid for $x \ne -3$, $x \ne -2$, and $x \ne 3$. A constant advertises no restrictions at all, so every one of them has to be carried along.

Factor all four quadratics.

$x^2-2x-3 = (x-3)(x+1), \quad x^2+4x+3 = (x+1)(x+3)$ $x^2+2x-3 = (x+3)(x-1), \quad x^2-4x+3 = (x-1)(x-3)$

The denominators vanish at $x = -1$, $x = -3$, $x = 1$, and $x = 3$, so those four values are excluded. Multiplying across, the top and the bottom hold exactly the same four factors.

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

The expression equals $1$, but only on its domain: at the four excluded values it has no value at all.

Factor, then write the polynomial divisor over $1$ and flip it.

$x^2+x-12 = (x+4)(x-3), \qquad x^2-1 = (x-1)(x+1)$

The denominator gives $x \ne 1$ and $x \ne -1$, and the divisor $x+4$ may not be zero, so $x \ne -4$.

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

Nothing in the answer contains $x+4$, so the restriction $x \ne -4$ survives only in your notes.

Multiplication and division run left to right, so factor everything and turn the division into multiplication by the reciprocal.

$x^2-4 = (x-2)(x+2), \quad x^2+3x = x(x+3)$ $x^2+3x-18 = (x+6)(x-3), \quad x^2+x-6 = (x+3)(x-2)$

Now write it as one big fraction and cancel the factors $x$, $x-2$, and $x+6$.

$\frac{(x-2)(x+2)(x+6)(x-3) \cdot x}{x(x+3)(x+3)(x-2)(x+6)} = \frac{(x+2)(x-3)}{(x+3)^2}$

The restrictions are $x \ne -6$, $x \ne -3$, $x \ne 0$, and $x \ne 2$.

Use the sum of cubes $a^3 + b^3 = (a+b)(a^2-ab+b^2)$ with $a = x$ and $b = 3$.

$x^3+27 = (x+3)(x^2-3x+9), \qquad x^2-9 = (x-3)(x+3)$

Multiply across and cancel $x^2-3x+9$ and one factor of $x+3$.

$\frac{(x+3)(x^2-3x+9)}{x^2-3x+9} \cdot \frac{x}{(x-3)(x+3)} = \frac{x}{x-3}$

The restrictions are $x \ne 3$ and $x \ne -3$. The quadratic $x^2-3x+9$ has discriminant $9 - 36 = -27 < 0$, so it is never zero for a real $x$ and adds no restriction.

The original carries three conditions: $x \ne -4$ from the dividend's denominator, $x \ne 2$ from the divisor's denominator, and $x \ne 1$ because the divisor $\frac{x-1}{x-2}$ is zero when $x = 1$.

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

The simplified form has the single denominator $x+4$, so it is undefined only at $x = -4$. At $x = 1$ it gives $-\tfrac{1}{5}$ and at $x = 2$ it gives $0$, both ordinary values, while the original expression exists at neither.

Factor both denominators and the divisor's numerator.

$x^2-25 = (x-5)(x+5), \qquad x^2-4x+3 = (x-1)(x-3)$

The dividend's denominator gives $x \ne 5$ and $x \ne -5$. The divisor's denominator gives $x \ne -5$. The divisor's numerator may not be zero, so $x \ne 1$ and $x \ne 3$.

$x \ne -5, \qquad x \ne 1, \qquad x \ne 3, \qquad x \ne 5$

The dividend's numerator $x+1$ is free to vanish, so $x = -1$ stays in the domain.

The first numerator is a perfect square.

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

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

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

The restrictions are $x \ne 3$, $x \ne -2$, and $x \ne 0$; only the last is visible in the answer.

Factor the cubic. The rational root theorem offers $\pm 1, \pm 2, \pm 3, \pm 6$, and $x = -1$ works, since $-1 - 4 - 1 + 6 = 0$. Synthetic division by $x+1$ leaves $x^2-5x+6$.

$x^3-4x^2+x+6 = (x+1)(x^2-5x+6) = (x+1)(x-2)(x-3)$

With $x^2-4 = (x-2)(x+2)$, multiply across and cancel $x-2$, $x+2$, and $x-3$.

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

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

Factor, and rewrite $2-x$ as $-(x-2)$ so the factors match.

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

Flip the divisor and multiply.

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

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

Factor and collect the three conditions.

$x^2-x-20 = (x-5)(x+4), \qquad x^2-4 = (x-2)(x+2)$

The dividend's denominator gives $x \ne -2$. The divisor's denominator $x^2-4$ gives $x \ne 2$ and $x \ne -2$. The divisor's numerator may not be zero, so $x \ne 5$.

$\frac{(x-5)(x+4)}{x+2} \cdot \frac{(x-2)(x+2)}{x-5} = (x+4)(x-2)$

The excluded values are $x = -2$, $x = 2$, and $x = 5$: three of them. The simplified answer is a polynomial and shows none of them.

Factor $x^2-1 = (x-1)(x+1)$ and collect the conditions from the original. The dividend's denominator gives $x \ne 1$ and $x \ne -1$. The divisor's denominator gives $x \ne 1$. The divisor's numerator $(x-4)^2$ is zero exactly at $x = 4$, so $x \ne 4$.

Now flip and multiply, canceling one factor of $x-1$.

$\frac{x+7}{(x-1)(x+1)} \cdot \frac{x-1}{(x-4)^2} = \frac{x+7}{(x+1)(x-4)^2}$

The answer displays $x \ne -1$ and $x \ne 4$, but $x \ne 1$ vanished with the canceled factor and has to be carried by hand.