Equivalent Rational Expressions — AP® Precalculus

Part 1 · Hook

Why this matters

A bakery prices custom cakes with a formula that is a quotient of two polynomials. Written one way, it shows where the price is undefined; rewritten by division, it shows the price per extra tier for very large orders. Same function, different questions answered.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. Factor $x^2 - x - 6$ .

$(x - 3)(x + 2)$
$(x + 3)(x - 2)$
$(x - 6)(x + 1)$
$(x - 3)(x - 2)$

Show the answer

Two numbers with product $-6$ and sum $-1$ : $-3$ and $2$ .

Correct: $(x - 3)(x + 2)$ :
$(x + 3)(x - 2)$ :
$(x - 6)(x + 1)$ :
$(x - 3)(x - 2)$ :

2. Where does $f(x)$ $= \frac{x - 1}{(x - 1)(x + 4)}$ have a hole?

$x = 1$
$x = -4$
$x = -1$
Nowhere

Show the answer

The factor $x - 1$ cancels, so the graph has a hole at $x = 1$ ; $x = -4$ is a vertical asymptote.

Correct: $x = 1$ :
$x = -4$ :
$x = -1$ :
Nowhere:

Part 4 · See it

See it first

Graph of r(x) = (x^2 - x - 2)/(x - 3): zeros at x = -1 and x = 2, a dashed vertical asymptote at x = 3, and the dashed slant asymptote y = x + 2 that the graph approaches at both ends. — One function, three forms: $r(x)$ $= \frac{x^2 - x - 2}{x - 3}$ $= \frac{(x - 2)(x + 1)}{x - 3}$ $= x$ $+ 2$ $+ \frac{4}{x - 3}$ . Factored form gives the zeros and the vertical asymptote; the divided form gives the slant asymptote $y = x + 2$ . LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Factor the numerator and the denominatorread zeros, holes, vertical asymptotes and the domain
Multiply out to standard formread the leading term and so the end behavior
Divide the numerator by the denominatorget quotient $+$ remainder over divisor, and the quotient is the slant or horizontal asymptote
Use a row of Pascal's triangleexpand $(a + b)^n$ without multiplying it out

Part 6 · Key ideas

Key ideas

Equivalent forms give the same output for every input in the domain; each form makes different features easy to read.
Polynomial long division writes dividend $=$ divisor $\cdot$ quotient $+$ remainder, so $\frac{p(x)}{d(x)}$ $= q(x)$ $+ \frac{r(x)}{d(x)}$ .
When the quotient is linear, the line $y = q(x)$ is a slant asymptote: the remainder term shrinks toward $0$ for large inputs.
Row $n$ of Pascal's triangle gives the coefficients of $(a + b)^n$ ; each entry is the sum of the two above it.

Part 7 · Misconception

A common mistake

The wrong idea: The remainder from dividing $p(x)$ by $x - 3$ can be ignored, so $\frac{p(x)}{x - 3}$ equals the quotient.

What actually happens: The remainder stays as $\frac{\text{remainder}}{x - 3}$ . It is small only for large inputs, which is why the quotient is an asymptote and not the function itself.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

The graph of $r(x)$ $= \frac{x^2 - x - 2}{x - 3}$

1. The graph shows $r$ . Which form of $r(x)$ shows its slant asymptote directly?

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

Show the answer

Long division gives $x^2 - x - 2$ $= (x - 3)(x + 2) + 4$ , so $r(x) = x + 2 + \frac{4}{x - 3}$ . As $x$ grows without bound, $\frac{4}{x - 3}$ shrinks toward $0$ and the graph approaches the line $y = x + 2$ .

$\frac{(x - 2)(x + 1)}{x - 3}$ : Factored form shows the zeros $2$ and $-1$ and the vertical asymptote, not the slant asymptote.
Correct: $x + 2 + \frac{4}{x - 3}$ : The quotient $x + 2$ is the slant asymptote, and the remainder term fades for large inputs.
$\frac{x^2 - x - 2}{x - 3}$ : This is the original quotient; the line it approaches is hidden until you divide.
$x - 3 + \frac{4}{x + 2}$ : This swaps the divisor and the quotient. Check it: at $x = 0$ it gives $-3 + 2 = -1$ , but $r(0) = \frac{-2}{-3} = \frac{2}{3}$ .

2. Divide to find the slant asymptote of $r(x)$ $= \frac{x^2 - x - 2}{x - 3}$ . Write the line.

Type an expression for $y =$ .

Show the answer

Divide $x^2 - x - 2$ by $x - 3$ : $x^2 \div x = x$ ; subtract $x(x - 3) = x^2 - 3x$ to leave $2x - 2$ ; $2x \div x = 2$ ; subtract $2(x - 3) = 2x - 6$ to leave the remainder $4$ . The quotient is $x + 2$ , so the slant asymptote is $y = x + 2$ .

Answer: $y =$ $x + 2$

3. Using the factored form $r(x)$ $= \frac{(x - 2)(x + 1)}{x - 3}$ , which values of $x$ are zeros of $r$ ? Select all that apply.

$x = 2$
$x = -1$
$x = 3$
$x = -2$

Show the answer

A zero of a rational function is a zero of the numerator that is in the domain: $x - 2 = 0$ and $x + 1 = 0$ give $x = 2$ and $x = -1$ , and neither makes the denominator $0$ .

Correct: $x = 2$ : $x - 2 = 0$ at $x = 2$ , and the denominator is $-1$ there.
Correct: $x = -1$ : $x + 1 = 0$ at $x = -1$ , and the denominator is $-4$ there.
$x = 3$ : $x = 3$ makes the denominator $0$ : it is the vertical asymptote, not a zero.
$x = -2$ : At $x = -2$ the numerator is $(-4)(-1) = 4$ , not $0$ ; this sign slip reads $x + 1$ as $x + 2$ .

4. Divide $x^3 + x^2 - 4x + 5$ by $x - 1$ . Write the quotient.

Type an expression for $q(x) =$ .

Show the answer

Synthetic division with $1$ : bring down $1$ ; $1 \cdot 1 + 1 = 2$ ; $2 \cdot 1 - 4 = -2$ ; $-2 \cdot 1 + 5 = 3$ . The quotient is $x^2 + 2x - 2$ with remainder $3$ , so $x^3$ $+ x^2 - 4x$ $+ 5$ $= (x - 1)(x^2 + 2x - 2)$ $+ 3$ .

Answer: $q(x) =$ $x^{2} + 2x - 2$

5. Which form of $g(x) = 2x^3 - 4x^2 - 6x$ shows its zeros and which shows its end behavior?

$2x(x - 3)(x + 1)$ shows the zeros; $2x^3 - 4x^2 - 6x$ shows the end behavior
$2x^3 - 4x^2 - 6x$ shows the zeros; $2x(x - 3)(x + 1)$ shows the end behavior
$2x(x - 3)(x + 1)$ shows both, and standard form shows neither
$2x(x + 3)(x - 1)$ shows the zeros; $2x^3 - 4x^2 - 6x$ shows the end behavior

Show the answer

Factor: $2x^3 - 4x^2 - 6x$ $= 2x(x^2 - 2x - 3)$ $= 2x(x - 3)(x + 1)$ , so the zeros are $0, 3, -1$ . The leading term $2x^3$ of standard form gives the end behavior.

Correct: $2x(x - 3)(x + 1)$ shows the zeros; $2x^3 - 4x^2 - 6x$ shows the end behavior: Each factor gives a zero; the leading term $2x^3$ gives the end behavior.
$2x^3 - 4x^2 - 6x$ shows the zeros; $2x(x - 3)(x + 1)$ shows the end behavior: Standard form hides the zeros; you must factor to see them.
$2x(x - 3)(x + 1)$ shows both, and standard form shows neither: Factored form can show end behavior after you multiply the leading parts, but standard form shows it at a glance, so it is not "neither".
$2x(x + 3)(x - 1)$ shows the zeros; $2x^3 - 4x^2 - 6x$ shows the end behavior: The signs are swapped: $(x + 3)(x - 1)$ $= x^2 + 2x - 3$ , not $x^2 - 2x - 3$ .

Table

Pascal's triangle, rows 0 to 6

Rows of Pascal's triangle: row n lists the coefficients of the expansion of (a + b) to the power n
Row $n$	Coefficients
0	1
1	1 1
2	1 2 1
3	1 3 3 1
4	1 4 6 4 1
5	1 5 10 10 5 1
6	1 6 15 20 15 6 1

6. Use the table. What is the coefficient of $x^2$ in the expansion of $(x + 3)^4$ ?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Row $4$ is $1, 4, 6, 4, 1$ . The $x^2$ term is $6x^2(3)^2 = 6 \cdot 9x^2 = 54x^2$ .

Answer: 54

Part 9 · Summary

Summary

Factored form shows zeros, holes, vertical asymptotes and the domain; standard form shows end behavior; division writes a rational function as quotient plus remainder over divisor, and a linear quotient is the slant asymptote. Pascal's triangle expands

(a + b)^n

.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

The graph of r(x)r(x)=x2−x−2x−3= \frac{x^2 - x - 2}{x - 3}

Pascal's triangle, rows 0 to 6

Summary

What comes next

Connections

The graph of $r(x)$ $= \frac{x^2 - x - 2}{x - 3}$