Simplifying Rational Expressions: Practice

Factor the numerator by grouping the first two terms and the last two.

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

The denominator is $(x-2)(x+2)$, zero at $x=2$ and $x=-2$, so both values are excluded before anything is canceled.

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

Both restrictions vanished from view, so both must be written down. Check at $x=0$: the original is $\frac{-4}{-4}=1$, and $x+1$ gives $1$.

The excluded values are the zeros of the denominator, so factor it.

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

Setting each factor to zero gives $x=-\tfrac{1}{2}$ and $x=3$, so those two are excluded. Add them.

$-\frac{1}{2} + 3 = \frac{5}{2}$

Vieta's formulas give the same total without factoring: the roots of $ax^2+bx+c$ sum to $-\frac{b}{a} = -\frac{-5}{2} = \frac{5}{2}$.

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

$2x^3-x^2-8x+4 = x^2(2x-1) - 4(2x-1) = (2x-1)(x^2-4)$

The denominator is $x^2-4 = (x-2)(x+2)$, zero at $x=2$ and $x=-2$, so both are excluded.

$\frac{(2x-1)(x^2-4)}{x^2-4} = 2x-1 \qquad (x \neq 2 \text{ and } x \neq -2)$

Check at $x=0$: the original is $\frac{4}{-4} = -1$, and $2(0)-1 = -1$ agrees.

Since $2^3-8=0$, the Factor Theorem says $x-2$ divides $x^3-8$, and dividing gives

$x^3-8 = (x-2)(x^2+2x+4).$

The only zero of the denominator is $x=2$, so that is the only excluded value. For every other $x$ the factor $x-2$ is nonzero and cancels, leaving $x^2+2x+4$.

At $x=2$ the left side is $\frac{0}{0}$, undefined, while the right side is $4+4+4=12$. So the statement is true for every real number except $x=2$. Nothing excludes $x=-2$: there the original is $\frac{-16}{-4}=4$, which matches $4-4+4=4$.

Factor the denominator by grouping.

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

The denominator is zero at $x=2$, $x=1$, and $x=-1$, so all three values are excluded.

It is tempting to keep only $x=2$, because the factors $x-1$ and $x+1$ also appear in the numerator and will cancel. They still make the original denominator zero, and the original is what defines the domain. Canceling later cannot un-forbid an input.

From the grouping above, the denominator is $(x-2)(x-1)(x+1)$, so the excluded values are $x=2$, $x=1$ and $x=-1$. The numerator is $(x-1)(x+1)$.

$\frac{(x-1)(x+1)}{(x-2)(x-1)(x+1)} = \frac{1}{x-2} \qquad (x \neq 2,\; x \neq 1,\; x \neq -1)$

Both canceled factors take their restrictions out of sight. The form $\frac{1}{x-2}$ is perfectly well defined at $x=1$ and $x=-1$, but the original is not, so all three restrictions belong in the answer.

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

If the quotient reduces to $x+3$ after canceling $x-4$, then the numerator must be the product of $x-4$ and $x+3$. Multiply them out.

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

Matching coefficients with $x^2+bx+c$ gives $b=-1$ and $c=-12$.

Check: $\frac{x^2-x-12}{x-4} = \frac{(x-4)(x+3)}{x-4} = x+3$ for $x \neq 4$, as required.

The expression is undefined exactly where the denominator is zero, so solve $x^2+4=0$.

$x^2 = -4$

No real number has a negative square, so the denominator is never zero and the expression is defined at every real number. The answer is zero values.

The zeros of the numerator, $x=2$ and $x=-2$, do not make the expression undefined; there it simply equals $\frac{0}{8}=0$.

Only a common factor cancels, so factor before deciding.

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

That one is a genuine cancellation, and it holds at every $x$ in the domain.

The others cancel terms, which is not a rule. In $\frac{x+7}{7}$ the $7$ on top is added, not multiplying, and at $x=1$ the value is $\frac{8}{7}$, not $1$. Likewise $\frac{x^2+9}{9}$ at $x=1$ is $\frac{10}{9}$, not $1$. And $x+3$ is not a factor of $x^2+9$, since $(-3)^2+9 = 18 \neq 0$; at $x=1$ the quotient is $\frac{10}{4}$, while $x+3=4$.

The denominator has $2$ as a root, since $2^3-8=0$, so by the Factor Theorem $x-2$ divides it, and dividing gives $x^3-8 = (x-2)(x^2+2x+4)$.

The quadratic $x^2+2x+4$ has discriminant $2^2-4(1)(4) = -12$, which is negative, so it has no real zeros. The only excluded value is therefore $x=2$.

Factor the numerator as $(3x+1)(x-2)$, which you can check by expanding: $3x^2-6x+x-2 = 3x^2-5x-2$.

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

Check at $x=1$: the original is $\frac{-4}{-7} = \frac{4}{7}$, and $\frac{4}{7}$ agrees.

First check whether $x=4$ is in the domain, by testing the denominator there.

$4^2-9 = 7 \neq 0$

It is nonzero, so $x=4$ is allowed. Now evaluate the numerator: $4^2-7(4)+12 = 16-28+12 = 0$.

$\frac{0}{7} = 0$

A zero on top is not a problem; only a zero on the bottom is. The expression simplifies to $\frac{x-4}{x+3}$ for $x \neq 3$ and $x \neq -3$, and that form also gives $\frac{0}{7} = 0$ at $x=4$.

For $x-5$ to cancel and leave $x+5$, the numerator must be the product $(x-5)(x+5)$.

$(x-5)(x+5) = x^2-25$

Matching this with $x^2-k$ gives $-k = -25$, so $k=25$.

Check: $\frac{x^2-25}{x-5} = \frac{(x-5)(x+5)}{x-5} = x+5$ for $x \neq 5$, exactly as required.