Multiplying and Dividing Rational Expressions: Additional Practice (Set 2)

Read the original, not the answer. Both fractions have denominator $x+5$, so $x \ne -5$. The divisor $\frac{x-4}{x+5}$ is zero exactly when its numerator is zero, and dividing by zero is undefined, so $x \ne 4$.

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

The simplified answer is a polynomial, defined at every real number, so it hides both restrictions. The original expression exists at neither $x = -5$ nor $x = 4$.

Factor first.

$2x^2+7x+3 = (2x+1)(x+3), \qquad x^2-2x-15 = (x-5)(x+3)$

Collect the conditions from the original: the dividend's denominator gives $x \ne 5$ and $x \ne -3$, the divisor's denominator gives $x \ne 5$, and the divisor's numerator may not be zero, so $2x + 1 \ne 0$ and $x \ne -\tfrac{1}{2}$.

$\frac{(2x+1)(x+3)}{(x-5)(x+3)} \cdot \frac{x-5}{2x+1} = 1$

The expression equals $1$, but only on its domain, so all three restrictions stay.

Simplifying gives $R(x) = x - 1$, which would suggest $R(2) = 1$. But $R$ is the original expression, and at $x = 2$ both of its denominators are zero.

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

That is undefined, so $x = 2$ is not in the domain of $R$ and $R(2)$ does not exist. The correct statement is $R(x) = x-1$ for $x \ne 2$ and $x \ne -1$, the second restriction coming from the divisor, which may not be zero.

Work left to right. Factor $x^2-1 = (x-1)(x+1)$ and $x^2+5x+6 = (x+2)(x+3)$, then multiply the first two.

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

Now divide by $\frac{x+1}{x+2}$, which means multiplying by its reciprocal.

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

The restrictions come from the original: $x \ne -2$ and $x \ne -3$ from the first denominator, $x \ne 1$ from the second, $x \ne -2$ from the divisor's denominator, and $x \ne -1$ because the divisor may not be zero.

Factor the quadratics.

$6x^2-x-2 = (3x-2)(2x+1), \qquad 4x^2-4x+1 = (2x-1)^2$

Multiply across, then cancel $3x-2$ and one factor of $2x-1$.

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

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

Factor, taking the common $x^2$ out of the first numerator.

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

Multiply across and cancel $x^2$ along with one factor of $x-2$.

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

The restrictions are $x \ne 2$ and $x \ne 0$; the answer shows only the first.

Simplify $f$ by factoring $x^2-4 = (x-2)(x+2)$ and cancelling.

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

But the cancelled factors leave restrictions behind: $f$ has denominators $x+3$ and $x-2$, so it is undefined at $x = -3$ and $x = 2$. The polynomial $g(x) = x+2$ is defined at every real number, in particular at those two.

So $f$ and $g$ have the same formula but different domains, which makes them different functions. The value $x = -2$ is a red herring: it only makes a numerator zero, and $f(-2) = 0$ exists.

Factor the difference of squares and the quadratic beneath it.

$9x^2-4 = (3x-2)(3x+2), \qquad 3x^2+x-2 = (3x-2)(x+1)$

Multiply across and cancel $3x-2$ and $x+1$.

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

The restrictions are $x \ne \tfrac{2}{3}$, $x \ne -1$, and $x \ne 0$.

The algebra is right, but two of the three conditions were never collected. The divisor's denominator gives $x + 1 \ne 0$, so $x \ne -1$. The divisor itself may not be zero, and it is zero exactly when its numerator vanishes.

$x + 5 = 0 \;\Rightarrow\; x = -5 \quad \text{is excluded}$

Both restrictions disappeared from the page: $x+1$ moved into a numerator when the divisor was flipped, and $x+5$ then cancelled. The complete answer is $\frac{x+1}{x-2}$ with $x \ne -5$, $x \ne -1$, and $x \ne 2$.

Use the difference of cubes $a^3-b^3 = (a-b)(a^2+ab+b^2)$ with $b = 1$, and factor the rest.

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

Multiply across and cancel $x^2+x+1$, $x-1$, and one factor of $x+1$.

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

The restrictions are $x \ne 1$ and $x \ne -1$. The quadratic $x^2+x+1$ has discriminant $1 - 4 = -3 < 0$, so it is never zero for a real $x$ and adds nothing.

Factor everything, pulling the $5$ out of the first numerator.

$5x-15 = 5(x-3), \quad x^2-x-6 = (x-3)(x+2), \quad x^2-4 = (x-2)(x+2)$

Flip the divisor and multiply, cancelling $x-3$, $x+2$, and the common factor $5$.

$\frac{5(x-3)}{(x-3)(x+2)} \cdot \frac{(x-2)(x+2)}{10} = \frac{5(x-2)}{10} = \frac{x-2}{2}$

The restrictions are $x \ne 3$, $x \ne -2$, and $x \ne 2$. The divisor's numerator is the constant $10$, which is never zero, so it adds no restriction of its own.

Factor all four quadratics.

$x^2-16 = (x-4)(x+4), \quad x^2+5x+6 = (x+2)(x+3)$ $x^2-8x+16 = (x-4)^2, \quad x^2-4 = (x-2)(x+2)$

Collect the conditions: the dividend's denominator gives $x \ne -2$ and $x \ne -3$, the divisor's denominator gives $x \ne 2$ and $x \ne -2$, and the divisor's numerator $(x-4)^2$ is zero at $x = 4$, so $x \ne 4$.

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

The answer shows $x \ne -3$ and $x \ne 4$; the restrictions $x \ne -2$ and $x \ne 2$ rode on cancelled factors and survive only in writing.