Simplifying Rational Expressions: Practice

A rational expression is a quotient of two polynomials, with the denominator not the zero polynomial.

The first two are quotients of polynomials, and $x^2+4x$ counts as well, because it can be written with denominator $1$.

$x^2+4x = \frac{x^2+4x}{1}$

The absolute value $|x|$ is not a polynomial (no sum of powers of $x$ with constant coefficients produces it), so $\dfrac{|x|+1}{x-2}$ is not a rational expression.

The domain of a rational expression is every real number that does not make the denominator zero, so solve the denominator equal to zero.

$x-7=0 \;\Longrightarrow\; x=7$

At $x=7$ the expression becomes $\frac{5}{0}$, which has no value. Every other real number is allowed, so the domain is all reals except $7$. The numerator $5$ never causes trouble.

Excluded values are the zeros of the denominator, and here the denominator is just $x$.

$x = 0$

At $x=0$ the expression reads $\frac{3}{0}$, which is undefined. The zero of the numerator, $x=-3$, is fine: there the expression equals $\frac{0}{-3}=0$, a perfectly good number.

The denominator is already factored, so set each factor equal to zero. A product is zero exactly when one of its factors is zero.

$x-4=0 \;\Longrightarrow\; x=4, \qquad x+1=0 \;\Longrightarrow\; x=-1$

So $4$ and $-1$ are excluded. The numerator's zero, $x=0$, stays in the domain: there the expression is $\frac{0}{-4}=0$.

The denominator $2x$ is zero at $x=0$, so $x=0$ is excluded before any cancelling happens.

Write the top and the bottom as products with the common factor $2x$ shown.

$\frac{6x^3}{2x} = \frac{(2x)(3x^2)}{(2x)(1)} = 3x^2 \qquad (x \neq 0)$

The cancellation is legal exactly because $2x \neq 0$ on the domain. The polynomial $3x^2$ is defined at $x=0$, while the original is not, so the restriction has to be written down.

The $5$ in the numerator is a term, not a factor of the whole numerator, so it cannot be cancelled against the $5$ underneath. Dividing a numerator by $5$ divides every term of it.

$\frac{x+5}{5} = \frac{x}{5} + \frac{5}{5} = \frac{x}{5} + 1$

One input settles it: at $x=5$ the expression is $\frac{10}{5}=2$, which is neither $x=5$ nor $x+1=6$. The numerator $x+5$ and the denominator $5$ share no common non-constant factor, so the expression is already in lowest terms.

Factor the denominator as a difference of squares and set each factor to zero.

$x^2-9 = (x-3)(x+3) = 0 \;\Longrightarrow\; x=3 \text{ or } x=-3$

Both values are excluded. The numerator's zero, $x=2$, is not excluded: there the expression equals $\frac{0}{-5}=0$.

Factor a constant out of the top and the bottom, then read the excluded value off the original denominator: $2x+4=0$ at $x=-2$.

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

The common factor $x+2$ is nonzero everywhere on the domain, which is what makes the cancellation legal. The constant $2$ is defined at $x=-2$, so the restriction has to be stated.

Both parts are already factored. The denominator is zero at $x=-2$ and $x=-7$, so those two values are excluded before anything is cancelled.

$\frac{(x-5)(x+2)}{(x+2)(x+7)} = \frac{x-5}{x+7} \qquad (x \neq -2 \text{ and } x \neq -7)$

Only the common factor $x+2$ may go. The $x$ inside $x-5$ and $x+7$ is a term of each, not a factor, so nothing further cancels and $\frac{-5}{7}$ is not the answer. Note that $x \neq -2$ has vanished from the simplified form, which is why it is written beside it.

Factor and cancel, keeping track of where the cancelled factor is nonzero.

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

At every $x \neq 5$ both sides are defined and equal. At $x=5$ the left side is $\frac{0}{0}$, which is undefined, while the right side is $10$, so the statement fails there. Cancelling preserves the value but not the domain.

Factor first, then read the excluded values off the original denominator: $(x-3)(x+3)$ is zero at $x=3$ and $x=-3$.

$\frac{x^2+7x+12}{x^2-9} = \frac{(x+3)(x+4)}{(x-3)(x+3)} = \frac{x+4}{x-3} \qquad (x \neq 3 \text{ and } x \neq -3)$

Both restrictions survive, even though only $x \neq 3$ is still visible in the simplified expression. Cancelling the factor $x+3$ removed the only reminder that $-3$ was forbidden, so it must be recorded by hand.

A rational expression is undefined exactly where its denominator is zero.

$x+6=0 \;\Longrightarrow\; x=-6$

At $x=-6$ the expression reads $\frac{3}{0}$, and no number multiplied by $0$ gives $3$, so there is no value to report. Everywhere else it is defined.