Simplifying Rational Expressions: Additional Practice (Set 1)

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

$3x^2-11x-4 = (3x+1)(x-4)$

Setting each factor to zero gives $x=-\tfrac{1}{3}$ and $x=4$. Multiply them.

$\left(-\frac{1}{3}\right)(4) = -\frac{4}{3}$

Vieta's formulas confirm it without factoring: the roots of $ax^2+bx+c$ multiply to $\frac{c}{a} = \frac{-4}{3}$.

Factor the numerator with two numbers multiplying to $-8$ and adding to $2$, namely $4$ and $-2$, and factor the denominator as a difference of squares.

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

The common factor $x-2$ is nonzero on the domain (which excludes $x=2$ and $x=-2$), so the cancellation is legal there.

Check at $x=0$: the original is $\frac{-8}{-4} = 2$, and $\frac{4}{2} = 2$ agrees.

The denominator is a difference of squares, since $4x^2 = (2x)^2$.

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

It is zero at $x=\tfrac{1}{2}$ and $x=-\tfrac{1}{2}$, so both values are excluded. The numerator factors as $(3x-2)(2x+1)$, which you can check by expanding: $6x^2+3x-4x-2 = 6x^2-x-2$.

$\frac{(3x-2)(2x+1)}{(2x-1)(2x+1)} = \frac{3x-2}{2x-1} \qquad \left(x \neq \tfrac{1}{2} \text{ and } x \neq -\tfrac{1}{2}\right)$

Cancelling $2x+1$ hid the restriction $x \neq -\tfrac{1}{2}$, so it is written beside the answer. Check at $x=0$: the original is $\frac{-2}{-1}=2$, and $\frac{-2}{-1}=2$ agrees.

The denominator is a perfect square, $(x+2)^2$, zero only at $x=-2$. So $x=-2$ is the one excluded value.

For the numerator, $(-2)^3+8 = 0$, so by the Factor Theorem $x+2$ is a factor of $x^3+8$. Dividing (synthetic division on the coefficients $1, 0, 0, 8$ with root $-2$) gives the quotient $x^2-2x+4$.

$\frac{x^3+8}{x^2+4x+4} = \frac{(x+2)(x^2-2x+4)}{(x+2)^2} = \frac{x^2-2x+4}{x+2} \qquad (x \neq -2)$

Check at $x=1$: the original is $\frac{9}{9}=1$, and $\frac{1-2+4}{3} = \frac{3}{3} = 1$ agrees. The sign of the middle term matters: $\frac{1+2+4}{3}$ would give $\frac{7}{3}$ instead.

Factor both quadratics. Expanding confirms each one.

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

The original denominator is zero when $2x-1=0$, that is $x=\tfrac{1}{2}$, and when $x-2=0$, that is $x=2$. Both values are excluded.

$\frac{(2x-1)(x+3)}{(2x-1)(x-2)} = \frac{x+3}{x-2} \qquad \left(x \neq \tfrac{1}{2} \text{ and } x \neq 2\right)$

Cancelling $2x-1$ took the restriction $x \neq \tfrac{1}{2}$ out of sight. Check at $x=0$: the original is $\frac{-3}{2}$, and $\frac{3}{-2} = -\frac{3}{2}$ agrees.

Factor the denominator as a difference of squares, then factor again.

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

The factor $x^2+4$ has no real zeros, since $x^2=-4$ has no real solution. So the only real zeros of the denominator come from $x-2$ and $x+2$, giving $x=2$ and $x=-2$.

That is two excluded values. The numerator $x^2+1$ never vanishes for real $x$ either, but that is irrelevant: only the denominator decides the domain.

The $x$ upstairs is a term, not a factor, so dividing the numerator by $x$ divides every term of it: $\frac{x+6}{x} = 1 + \frac{6}{x}$, which is not $6$ in general. Still, the two can coincide at a single input, so test the candidates.

$x = \frac{6}{5}: \quad \frac{\frac{6}{5}+6}{\frac{6}{5}} = \frac{\frac{36}{5}}{\frac{6}{5}} = 6$

The others fail: at $x=1$ the value is $\frac{7}{1} = 7$, at $x=6$ it is $\frac{12}{6} = 2$, and at $x=0$ the expression is undefined.

One lucky input is not a proof. A cancellation has to work at every point of the domain, and this one works at only one.

First check that $x=0$ is in the domain: the denominator there is $0-4 = -4$, which is not zero, so the value exists.

Substitute directly.

$\frac{0-0+4}{0-4} = \frac{4}{-4} = -1$

The simplified form gives the same answer. Factoring, $\frac{(x-2)^2}{(x-2)(x+2)} = \frac{x-2}{x+2}$ for $x \neq 2$ and $x \neq -2$, and at $x=0$ that is $\frac{-2}{2} = -1$.

Excluded values come only from the denominator, so solve each denominator equal to zero.

$x^2+2 = 0 \;\Longrightarrow\; x^2 = -2$

No real number has a negative square, so $x^2+2$ is never zero and $\frac{x^2-1}{x^2+2}$ excludes nothing.

The others all have real zeros downstairs: $x^2-1 = (x-1)(x+1)$ vanishes at $\pm 1$, $x^2-2x = x(x-2)$ vanishes at $0$ and $2$, and $x^2+2x+1 = (x+1)^2$ vanishes at $-1$.

The numerator is a difference of squares in disguise, with $(x+1)$ playing the role of the first quantity.

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

The denominator is the perfect square $(x+3)^2$, which is zero only at $x=-3$, so that is the only excluded value.

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

Nothing excludes $x=1$: there the original is $\frac{4-4}{16} = 0$. Check at $x=0$: the original is $\frac{1-4}{9} = -\frac{1}{3}$, and $\frac{-1}{3}$ agrees.

The denominator is $x^2-1 = (x-1)(x+1)$. For the quotient to reduce to $\frac{x+5}{x+1}$, the factor $x-1$ must cancel, so the numerator has to be $(x-1)(x+5)$.

$(x-1)(x+5) = x^2+5x-x-5 = x^2+4x-5$

Matching coefficients with $x^2+ax+b$ gives $a=4$ and $b=-5$.

Check: $\frac{x^2+4x-5}{x^2-1} = \frac{(x-1)(x+5)}{(x-1)(x+1)} = \frac{x+5}{x+1}$ for $x \neq 1$ and $x \neq -1$, as required.

Group the denominator in pairs.

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

The factor $x^2+1$ has no real zeros, so the only excluded value is $x=-1$.

The numerator is a difference of squares, twice.

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

Divide out the common factors $x+1$ and $x^2+1$, both nonzero on the domain.

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

Nothing excludes $x=1$: the denominator there is $1+1+1+1 = 4 \neq 0$. Check at $x=0$: the original is $\frac{-1}{1} = -1$, and $0-1 = -1$ agrees.