Simplifying Rational Expressions: Additional Practice (Set 2)

Treat $x^4-16$ as a difference of squares, since $x^4 = (x^2)^2$ and $16 = 4^2$.

$x^4-16 = (x^2-4)(x^2+4)$

The denominator $x^2-4$ is zero at $x=2$ and $x=-2$, so those are the excluded values. The common factor $x^2-4$ is nonzero everywhere else, so it divides out.

$\frac{(x^2-4)(x^2+4)}{x^2-4} = x^2+4 \qquad (x \neq 2 \text{ and } x \neq -2)$

The polynomial $x^2+4$ is defined at $\pm 2$, so the restrictions travel with it. Note that $x^2+4$ does not factor further over the reals, since $x^2=-4$ has no real solution.

Factor both parts. The numerator is a difference of squares, and the denominator has a common factor of $x^2$.

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

The original denominator is zero at $x=0$ and at $x=-3$, so both values are excluded. Divide out the common factor $x+3$.

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

The restriction $x \neq -3$ is invisible in the answer, so it is written beside it. Check at $x=1$: the original is $\frac{-8}{4} = -2$, and $\frac{-2}{1} = -2$ agrees.

The denominator is a perfect square, $x^2-6x+9 = (x-3)^2$, which is zero only at $x=3$. So $x=3$ is the one and only excluded value; $x=-3$ is in the domain, where the expression equals $\frac{0}{36}=0$.

The numerator is a difference of squares whose factor $3-x$ is the opposite of $x-3$.

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

Divide out one copy of the common factor $x-3$.

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

Check at $x=0$: the original is $\frac{9}{9}=1$, and $-\frac{3}{-3} = 1$ agrees. Losing the minus sign would give $-1$.

Group the numerator in pairs and pull out the common factor from each pair.

$x^3+2x^2-9x-18 = x^2(x+2) - 9(x+2) = (x+2)(x^2-9) = (x+2)(x-3)(x+3)$

The denominator factors as $(x+2)(x+3)$, which is zero at $x=-2$ and $x=-3$, so both values are excluded.

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

Check at $x=0$: the original is $\frac{-18}{6} = -3$, and $0-3 = -3$ agrees.

Factor the denominator by grouping.

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

The factor $x^2+1$ has no real zeros, so the only real zero of the denominator is $x=2$, and that is the only excluded value.

The numerator factors as $(x+3)(x-2)$, so the common factor $x-2$ divides out.

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

Check at $x=0$: the original is $\frac{-6}{-2}=3$, and $\frac{3}{1}=3$ agrees.

Factor each one and look for a common factor.

$\frac{x^2+4x+4}{x^2-4} = \frac{(x+2)^2}{(x-2)(x+2)} = \frac{x+2}{x-2} \qquad (x \neq 2 \text{ and } x \neq -2)$

That one has the common factor $x+2$, so it is not in lowest terms.

The others are reduced already. By the Factor Theorem, $x+2$ is not a factor of $x^2+4$, since $(-2)^2+4 = 8 \neq 0$. The polynomial $x^2+1$ has no real zeros, so it shares no linear factor with $x^2-1 = (x-1)(x+1)$. And $x+5$ and $x-5$ are different linear polynomials with no factor in common.

Factor the numerator with two numbers multiplying to $-15$ and adding to $-2$, namely $-5$ and $3$, and factor the denominator as a difference of squares.

$\frac{x^2-2x-15}{x^2-25} = \frac{(x-5)(x+3)}{(x-5)(x+5)} = \frac{x+3}{x+5} \qquad (x \neq 5 \text{ and } x \neq -5)$

The original denominator is zero at $x=5$ and $x=-5$, so both values are excluded. Cancelling $x-5$ removed every trace of the restriction $x \neq 5$, so it has to be stated.

Check at $x=0$: the original is $\frac{-15}{-25} = \frac{3}{5}$, and $\frac{3}{5}$ agrees.

Factor both parts of the left-hand side. The denominator is a perfect square.

$\frac{x^2-16}{x^2-8x+16} = \frac{(x-4)(x+4)}{(x-4)^2} = \frac{x+4}{x-4} \qquad (x \neq 4)$

The only zero of the original denominator is $x=4$, so that is the only excluded value. Nothing forbids $x=-4$: there the left side is $\frac{0}{64} = 0$ and the right side is $\frac{0}{-8}=0$, so they agree.

This time the surviving factor $x-4$ keeps the restriction visible, so both sides have the same domain and the equality holds everywhere on it.

Test each candidate for a common factor.

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

Each of those three has a common factor to divide out. For $\frac{x^2+x+1}{x+1}$, the Factor Theorem decides it: $x+1$ is a factor of $x^2+x+1$ only if $x=-1$ is a root, and $(-1)^2+(-1)+1 = 1 \neq 0$. There is no common non-constant factor, so $\frac{x^2+x+1}{x+1}$ is in lowest terms.

Factor the numerator, and read the excluded value off the original denominator: $x-2$ is zero at $x=2$.

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

So the expression is not in lowest terms, and it does agree with $x$ at every input in its domain. But it is not the same object as $x$: the polynomial $x$ returns $2$ at $x=2$, while the original there is $\frac{0}{0}$, which is undefined.

Cancelling preserves the value wherever both sides are defined, and it does not preserve the domain.

Factor a constant out of each part. The denominator is zero only at $x=2$, so that is the only excluded value.

$\frac{2x^2-8}{4-2x} = \frac{2(x-2)(x+2)}{-2(x-2)}$

Divide out the common factor $x-2$, which is nonzero on the domain, and reduce the constants.

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

Check at $x=0$: the original is $\frac{-8}{4} = -2$, and $-(0+2) = -2$ agrees. Nothing excludes $x=-2$, where the original is $\frac{0}{8}=0$.

Group the numerator in pairs.

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

The denominator factors as $(x-3)(x+2)$, which is zero at $x=3$ and $x=-2$, so both are excluded.

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

Check at $x=0$: the original is $\frac{12}{-6} = -2$, and $0-2 = -2$ agrees. Both restrictions were cancelled out of sight, so both are written down.