Simplifying Rational Expressions: Practice

Factor both parts. The numerator needs two numbers multiplying to $-6$ and adding to $-1$, namely $-3$ and $2$; the denominator is a difference of squares.

$\frac{x^2-x-6}{x^2-9} = \frac{(x-3)(x+2)}{(x-3)(x+3)} = \frac{x+2}{x+3} \qquad (x \neq 3 \text{ and } x \neq -3)$

The original denominator is zero at $x=3$ and $x=-3$, so both values are excluded. Only $x \neq -3$ is still visible in the simplified form, so $x \neq 3$ has to be carried along by hand.

Factor the denominator completely. Take out the common factor $x$ first, and what remains is a difference of squares.

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

A product is zero exactly when one of its factors is zero, so the denominator vanishes at $x=0$, $x=2$, and $x=-2$. All three are excluded. The numerator's zero, $x=-1$, is not excluded: there the expression equals $\frac{0}{3}=0$.

Take the common factor $3$ out of the numerator, which leaves a difference of squares, and factor the denominator into two numbers multiplying to $-6$ and adding to $1$.

$\frac{3x^2-12}{x^2+x-6} = \frac{3(x-2)(x+2)}{(x+3)(x-2)} = \frac{3(x+2)}{x+3} \qquad (x \neq -3 \text{ and } x \neq 2)$

The original denominator is zero at $x=-3$ and $x=2$, so both are excluded, and only $x \neq -3$ survives visibly. The factor $3$ is not cancelled by anything: there is no $3$ in the denominator to divide it out.

Check at $x=0$: the original is $\frac{-12}{-6}=2$, and $\frac{3(2)}{3}=2$ agrees.

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

$\frac{x^2-6x+9}{x^2-9} = \frac{(x-3)^2}{(x-3)(x+3)} = \frac{x-3}{x+3} \qquad (x \neq 3 \text{ and } x \neq -3)$

One copy of $x-3$ cancels, and one copy stays upstairs. The original denominator is zero at $x=3$ and $x=-3$, so both are excluded, and $x \neq 3$ is no longer visible in the answer: $\frac{x-3}{x+3}$ is defined at $x=3$ (it equals $0$ there) while the original is not.

The numerator is a perfect square, so factor and cancel one copy of $x+3$.

$\frac{x^2+6x+9}{x+3} = \frac{(x+3)^2}{x+3} = x+3 \qquad (x \neq -3)$

At every $x \neq -3$ both sides are defined and equal. At $x=-3$ the left side is $\frac{0}{0}$, which is undefined, while the right side is $0$. An undefined quantity is not equal to $0$, so the statement fails there and holds everywhere else.

The denominator factors as $(x-3)(x+3)$, so the excluded values are $x=3$ and $x=-3$.

For the numerator, note that $3^3-27 = 0$, so by the Factor Theorem $x-3$ is a factor of $x^3-27$. Dividing (synthetic division on the coefficients $1, 0, 0, -27$) gives the quotient $x^2+3x+9$.

$\frac{x^3-27}{x^2-9} = \frac{(x-3)(x^2+3x+9)}{(x-3)(x+3)} = \frac{x^2+3x+9}{x+3} \qquad (x \neq 3 \text{ and } x \neq -3)$

Check at $x=1$: the original is $\frac{-26}{-8} = \frac{13}{4}$, and $\frac{1+3+9}{4} = \frac{13}{4}$ agrees.

The denominator is zero at $x=4$, so that value is excluded. The two parts are opposites of each other.

$4-x = -(x-4)$

Substitute that in and divide out the common factor $x-4$, which is nonzero on the domain.

$\frac{x-4}{-(x-4)} = \frac{1}{-1} = -1 \qquad (x \neq 4)$

Check at $x=5$: the original is $\frac{1}{-1} = -1$. The answer is a constant, but only on the domain: at $x=4$ the original is $\frac{0}{0}$, not $-1$.

Excluded values come from the denominator, so solve $x^2+1=0$.

$x^2 = -1$

No real number has a negative square, so the denominator is never zero and nothing is excluded. The domain is every real number.

The zeros of the numerator, $x=1$ and $x=-1$, are in the domain: there the expression equals $\frac{0}{2}=0$.

Factor the constants out first, then factor what remains.

$\frac{6x^2-6}{3x^2-3x} = \frac{6(x-1)(x+1)}{3x(x-1)}$

The original denominator $3x(x-1)$ is zero at $x=0$ and $x=1$, so both are excluded. Divide out the common factor $x-1$ and reduce $\frac{6}{3}$ to $2$.

$\frac{6(x-1)(x+1)}{3x(x-1)} = \frac{2(x+1)}{x} \qquad (x \neq 0 \text{ and } x \neq 1)$

Check at $x=2$: the original is $\frac{18}{6}=3$, and $\frac{2(3)}{2}=3$ agrees. The restriction $x \neq 1$ is invisible in the answer, so it must be written down.

Factor both trinomials. For the numerator, $5$ and $-2$ multiply to $-10$ and add to $3$; for the denominator, $3$ and $-2$ multiply to $-6$ and add to $1$.

$\frac{x^2+3x-10}{x^2+x-6} = \frac{(x+5)(x-2)}{(x+3)(x-2)} = \frac{x+5}{x+3} \qquad (x \neq -3 \text{ and } x \neq 2)$

The original denominator is zero at $x=-3$ and $x=2$, so both values are excluded. Cancelling $x-2$ hid the second restriction, so it is recorded beside the answer.

Check at $x=0$: the original is $\frac{-10}{-6} = \frac{5}{3}$, and $\frac{5}{3}$ agrees.

Factor and cancel to see how far the agreement goes.

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

At every input other than $3$ the two agree. At $x=3$ the first is $\frac{0}{0}$, which has no value, while $x+3$ returns $6$. So they are not the same object: cancelling preserves the value wherever both sides are defined, but it does not preserve the domain.

The value $x=-3$ is a zero of the numerator and is in the domain of both, where each equals $0$.

Factor the denominator completely: take out the common factor $x$, and what remains is a difference of squares.

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

The product is zero exactly when one of the three factors is zero, so the denominator vanishes at $x=0$, $x=1$, and $x=-1$. That is three excluded values.

The numerator's zero, $x=-2$, is not excluded, since a zero on top is a perfectly good value.