Rational expressions: Algebra 2 Regents Practice

Simplify the rational expression $\frac{x^2-9}{x^2+7x+12}$.

• (A) $\frac{x+3}{x-4}$
• (B) $\frac{-9}{7x+12}$
• (C) $-\frac{3}{4}$
• (D) $\frac{x-3}{x+4}$ ✓

To simplify a rational expression, factor the numerator and denominator fully and cancel common factors; you may cancel factors but never individual terms.

Express the product in simplest form: $\frac{x^2-4}{x+5}\cdot\frac{2x+10}{x-2}$.

• (A) $x+2$
• (B) $2x-4$
• (C) $\frac{2x+4}{x+5}$
• (D) $2x+4$ ✓

When multiplying rational expressions, factor everything first, then cancel any factor shared by a numerator and a denominator before multiplying across.

For which value(s) of $x$ is the expression $\frac{x+5}{x^2-x-6}$ undefined?

• (A) $x=-3$ and $x=2$
• (B) $x=-5$
• (C) $x=3$ and $x=-2$ ✓
• (D) $x=3,\ x=-2,\ x=-5$

A rational expression is undefined wherever its denominator equals zero, so set only the denominator equal to zero to find the excluded values.

For which value(s) of $x$ is the expression $\dfrac{x+5}{x^2-9}$ undefined?

• (A) $x=-5$
• (B) $x=9$ and $x=-9$
• (C) $x=3$
• (D) $x=3$ and $x=-3$ ✓

A rational expression is undefined exactly where its denominator equals zero, regardless of the numerator.

For all values of $x$ for which the expression is defined, $\dfrac{x^2-9}{x^2-x-6}$ is equivalent to

• (A) $\dfrac{x+3}{x-2}$
• (B) $\dfrac{x+3}{x+2}$ ✓
• (C) $\dfrac{3}{2}$
• (D) $\dfrac{x-3}{x+2}$

Factor the numerator as a difference of squares and factor the denominator trinomial, then cancel the common factor $x-3$. Cancelling individual terms instead of factors leads to the wrong distractors.

Solve algebraically for all values of $x$: $$\frac{x+2}{x-1} = \frac{12}{x^2-1} + 1$$ Justify why any excluded values must be rejected.

Clearing the fractions with the common denominator $(x-1)(x+1)$ turns the rational equation into a linear one because the squared terms cancel. The single solution $x = 3$ must be checked against the excluded values $x = \pm 1$; since it is neither, it is valid.

For what value of $x$ is the expression $\dfrac{x+2}{x-3}$ undefined?

• (A) $x = -2$
• (B) $x = 2$
• (C) $x = 0$
• (D) $x = 3$ ✓

A rational expression is undefined exactly where the denominator equals zero.

Express the quotient in simplest form: $\frac{2x^2+6x}{x^2-9}\div\frac{x^2+3x}{x-3}$.

• (A) $\frac{2x}{x+3}$
• (B) $\frac{x+3}{2}$
• (C) $\frac{2}{x+3}$ ✓
• (D) $\frac{2x^2(x+3)}{(x-3)^2}$

Dividing by a rational expression means multiplying by its reciprocal; flip the divisor, then factor and cancel common factors.

Write $\frac{3}{x+2}+\frac{2}{x-1}$ as a single rational expression in simplest form.

• (A) $\frac{5}{2x+1}$
• (B) $\frac{5x+4}{(x+2)(x-1)}$
• (C) $\frac{3x-1}{(x+2)(x-1)}$
• (D) $\frac{5x+1}{(x+2)(x-1)}$ ✓

To add rational expressions, rewrite each one over the common denominator (the product of the distinct factors), then combine the numerators and simplify.

Solve for $x$: $\frac{5}{x-1}=\frac{3}{x+1}$.

• (A) $x=-4$ ✓
• (B) $x=-1$
• (C) $x=-8$
• (D) $x=4$

A proportion is solved by cross-multiplying; distribute carefully and confirm the solution does not make any denominator zero.

Simplify the complex fraction $\dfrac{\frac{1}{x}-\frac{1}{3}}{x-3}$.

• (A) $\dfrac{1}{3x}$
• (B) $\dfrac{3-x}{3x(x-3)}$
• (C) $-\dfrac{3}{x}$
• (D) $-\dfrac{1}{3x}$ ✓

A complex fraction is simplified by combining the top into a single fraction, then multiplying by the reciprocal of the bottom and cancelling common factors.

Solve for $x$: $\dfrac{x}{x-4}=\dfrac{4}{x-4}+2$.

• (A) No solution ✓
• (B) $x=4$
• (C) $x=2$
• (D) $x=-4$

Solutions that make an original denominator zero must be rejected as extraneous; here the only candidate is excluded, so there is no solution.

Simplify: $\dfrac{x^2-25}{x^2+8x+15}\div\dfrac{x-5}{x+1}$.

• (A) $\dfrac{x-5}{x+3}$
• (B) $\dfrac{x+3}{x+1}$
• (C) $\dfrac{x+1}{x+3}$ ✓
• (D) $\dfrac{(x-5)^2}{(x+3)(x+1)}$

To divide rational expressions, factor everything, multiply by the reciprocal of the divisor, then cancel common factors.

Write $\dfrac{x^2+3x+5}{x+1}$ in the form $q(x)+\dfrac{r}{x+1}$, where $r$ is a constant.

• (A) $x+2+\dfrac{3}{x+1}$ ✓
• (B) $x+3+\dfrac{2}{x+1}$
• (C) $x+4+\dfrac{1}{x+1}$
• (D) $x+2+\dfrac{5}{x+1}$

Polynomial long division rewrites an improper rational expression as a polynomial quotient plus a proper remainder term over the divisor.

For all values of $x$ for which the expression is defined, $\dfrac{x^2 - 9}{x^2 + x - 12}$ is equivalent to

• (A) $\dfrac{x+3}{x-4}$
• (B) $\dfrac{x-3}{x+4}$
• (C) $\dfrac{3}{4}$
• (D) $\dfrac{x+3}{x+4}$ ✓

Factoring the numerator as $(x-3)(x+3)$ and the denominator as $(x+4)(x-3)$ reveals the common factor $x-3$, which divides out where the expression is defined, leaving $\frac{x+3}{x+4}$. Canceling individual terms rather than factors leads to the incorrect distractors.

Solve algebraically for $x$: $\dfrac{x}{x-2} + \dfrac{2}{x+3} = 1$

Clearing the fractions by multiplying through by $(x-2)(x+3)$ turns the equation into a polynomial equation. The $x^2$ terms cancel, leaving a linear equation with solution $x=-\tfrac{1}{2}$, which must be checked against the restrictions $x \ne 2$ and $x \ne -3$; it satisfies both.

For all values of $x$ where the expression is defined, $\dfrac{x^2 - 9}{x^2 + x - 6}$ is equivalent to

• (A) $\dfrac{9}{x - 6}$
• (B) $\dfrac{x - 3}{x - 2}$ ✓
• (C) $\dfrac{x - 3}{x + 2}$
• (D) $\dfrac{x + 3}{x - 2}$

Factor both the numerator and denominator, then cancel the common factor $(x + 3)$. The key idea is that only common factors, not individual terms, can be cancelled, which rules out distractors like $\frac{9}{x-6}$.

Simplify completely: $\dfrac{x^2-9}{x^2+5x+6} \div \dfrac{x^2-3x}{x+2}$. State the simplified expression in lowest terms.

Rewriting the division as multiplication by the reciprocal and factoring completely reveals common factors $(x-3)$, $(x+3)$, and $(x+2)$ in numerator and denominator. After canceling, only $1/x$ remains.

For all values of $x$ for which the expression is defined, $\dfrac{x^2-4}{x^2+3x} \div \dfrac{x-2}{x+3}$ is equivalent to

• (A) $\dfrac{x+2}{x+3}$
• (B) $\dfrac{(x-2)^2(x+2)}{x(x+3)^2}$
• (C) $\dfrac{x-2}{x}$
• (D) $\dfrac{x+2}{x}$ ✓

To divide rational expressions, multiply by the reciprocal of the second fraction. After factoring, $x^2 - 4 = (x-2)(x+2)$ and $x^2 + 3x = x(x+3)$, the factors $(x-2)$ and $(x+3)$ cancel, leaving $\frac{x+2}{x}$. Multiplying instead of flipping the divisor produces the last distractor.

Simplify completely, for all values of $x$ for which the expression is defined: $\dfrac{x^2-9}{x^2+5x+6} \div \dfrac{x^2-3x}{x+2}$

Rewriting the division as multiplication by the reciprocal and factoring each polynomial reveals common factors $(x-3)$, $(x+3)$, and $(x+2)$ in the numerator and denominator. After canceling, only $\frac{1}{x}$ remains.

For all values of $x$ for which the expression is defined, $\dfrac{x^2-9}{x^2+5x+6}$ is equivalent to

• (A) $\dfrac{x-3}{x+2}$ ✓
• (B) $\dfrac{x+3}{x+2}$
• (C) $\dfrac{x+3}{x-2}$
• (D) $\dfrac{x-3}{x-2}$

Factor both the numerator and denominator: the numerator is a difference of squares, $(x-3)(x+3)$, and the denominator factors as $(x+2)(x+3)$. Cancelling the shared factor $(x+3)$ leaves $\frac{x-3}{x+2}$. The distractors come from cancelling the wrong factor or sign errors while factoring.

For all values of $x$ where the expression is defined, which expression is equivalent to $\dfrac{x^2-9}{x^2+x-6}$?

• (A) $\dfrac{x+3}{x+2}$
• (B) $\dfrac{x+3}{x-2}$
• (C) $\dfrac{x-3}{x+2}$
• (D) $\dfrac{x-3}{x-2}$ ✓

Factoring gives $\frac{(x-3)(x+3)}{(x+3)(x-2)}$. The common factor $x+3$ cancels, leaving $\frac{x-3}{x-2}$. The distractors come from factoring errors or canceling the wrong factor.

Solve algebraically for all values of $x$: $\dfrac{1}{x} + \dfrac{1}{x+2} = \dfrac{3}{4}$

Clearing denominators by multiplying through by $4x(x+2)$ converts the rational equation into the quadratic $3x^2-2x-8=0$, which factors as $(3x+4)(x-2)$. Both candidate solutions avoid the excluded values $x=0$ and $x=-2$, so both are valid.

Solve the equation $\dfrac{x^2}{x-3} = \dfrac{9}{x-3}$ algebraically for all values of $x$. Justify why any value you reject is not a solution.

Clearing the common denominator gives $x^2 = 9$, but multiplying by $x-3$ can introduce extraneous solutions. Since $x = 3$ makes the original denominators zero, it must be rejected, leaving $x = -3$ as the only solution.