1.11 - Equivalent Representations of Polynomial and Rational Expressions

Overview

The same polynomial or rational function can be written in more than one way, and each form makes different features easy to see. Factored form shows zeros, holes, and vertical asymptotes. Standard form shows end behavior. This topic is about picking the right form for the question, plus two tools for rewriting: polynomial long division (which also finds slant asymptotes) and the binomial theorem.

One function, several forms

Rewriting an expression doesn't change the function. The graph, the zeros, and every output stay the same. What changes is which features you can read off without extra work.

Form	Example	What it shows quickly
Factored polynomial	$f(x) = -2(x-1)^2(x+3)$	Real zeros, $x$ -intercepts, multiplicities
Standard polynomial	$f(x) = -2x^3 - 2x^2 + 10x - 6$	Degree, leading term, end behavior, $y$ -intercept
Factored rational	$r(x) = \dfrac{(x-2)(x+3)}{(x-2)(x+1)}$	Domain, zeros, holes, vertical asymptotes
Quotient plus remainder	$h(x) = x + 2 + \dfrac{1}{x-1}$	Slant asymptote, and which side of it the graph is on

Factored form: Zeros, holes, asymptotes, domain, and range

A product equals zero exactly when one of its factors equals zero, so factored form shows the real zeros right away. For a rational function

r(x) = \dfrac{p(x)}{q(x)}

, factor the numerator and the denominator completely, then read the features from the factors:

Domain: every real number except the zeros of the original denominator. Find these before you cancel anything.

Holes: a factor $(x - c)$ in both the numerator and the denominator (at least as many times on top as on the bottom) cancels and leaves a hole at $x = c$ . Plug $c$ into the simplified expression to get the hole's height.

Vertical asymptotes: zeros of the denominator that are left after canceling.

Zeros and x-intercepts: zeros of the numerator that are left after canceling and are in the domain.

Range: look for outputs the function never reaches, such as the height of a hole or a horizontal asymptote the graph never crosses.

Example 1: Reading a rational function from factored form

Let

r(x) = \dfrac{x^2 + x - 6}{x^2 - x - 2}

. Find the domain, hole, vertical asymptote, zero,

y

-intercept, horizontal asymptote, and range.

Step 1: Factor the numerator and the denominator.

r(x) = \frac{(x-2)(x+3)}{(x-2)(x+1)}

Step 2: Domain. The original denominator is zero at

x = 2

and

x = -1

, so the domain is all real numbers except

2

and

-1

Step 3: Hole. The factor

(x - 2)

cancels, so there's a hole at

x = 2

. Its height comes from the simplified expression

\dfrac{x+3}{x+1}

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

So the hole is at

\left(2, \frac{5}{3}\right)

Step 4: Vertical asymptote. After canceling,

x + 1

is still in the denominator, so

x = -1

is a vertical asymptote.

Step 5: Zero. After canceling,

x + 3

is still in the numerator, so

x = -3

is a zero and

(-3, 0)

is an

x

-intercept.

Step 6: The

y

-intercept is

r(0) = \dfrac{-6}{-2} = 3

, so the graph crosses the

y

-axis at

(0, 3)

Step 7: Horizontal asymptote. This one comes from standard form (next section). The numerator and denominator both have degree 2 and leading coefficient 1, so the horizontal asymptote is

y = \frac{1}{1} = 1

Step 8: Range. Setting

\dfrac{x+3}{x+1} = 1

gives

x + 3 = x + 1

, or

3 = 1

, which is impossible. So the graph never reaches

y = 1

. Setting

\dfrac{x+3}{x+1} = \dfrac{5}{3}

gives only

x = 2

, which is the hole, so

\frac{5}{3}

is never an output either.

Answer: The domain is all real numbers except

-1

and

2

. There's a hole at

\left(2, \frac{5}{3}\right)

, a vertical asymptote at

x = -1

, a zero at

x = -3

, a

y

-intercept of

3

, and a horizontal asymptote at

y = 1

. The range is all real numbers except

1

and

\frac{5}{3}

1.11 - Equivalent Representations of Polynomial and Rational Expressions

Standard form: End behavior

Standard form lists the terms from the highest power down. The first term, the leading term, controls the end behavior, because when

|x|

is large it's much bigger than all the other terms combined.

Polynomials: as $x \to \pm\infty$ , $p(x)$ behaves like its leading term $a_n x^n$ . Whether $n$ is even or odd, and the sign of $a_n$ , decide where each end of the graph goes.

Rational functions: compare the leading terms of the numerator and the denominator. If the denominator has the higher degree, $y = 0$ is the horizontal asymptote. If the degrees are equal, the horizontal asymptote is $y = \dfrac{\text{leading coefficient of the numerator}}{\text{leading coefficient of the denominator}}$ . If the numerator has the higher degree, there's no horizontal asymptote.

Factored form hides the leading term. To find it without expanding everything, multiply only the leading terms of the factors. For

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

, that's

-2 \cdot x^2 \cdot x = -2x^3

Example 2: Two forms of the same polynomial

The polynomial

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

expands to

f(x) = -2x^3 - 2x^2 + 10x - 6

. Use whichever form makes each question quick.

Zeros: use factored form. The zeros are $x = 1$ (multiplicity 2, so the graph touches the $x$ -axis and turns around) and $x = -3$ (multiplicity 1, so the graph crosses the $x$ -axis).

Vertical intercept: use standard form. The constant term is $-6$ , so the graph passes through $(0, -6)$ .

End behavior: use standard form. The leading term is $-2x^3$ , which has odd degree and a negative coefficient, so the left end goes up and the right end goes down:

\lim_{x \to -\infty} f(x) = \infty \qquad \text{and} \qquad \lim_{x \to \infty} f(x) = -\infty

Using different forms to answer questions in context

In a context problem, different parts of the question often call for different forms of the same function. Rewrite the function into whichever form answers each part fastest.

Example 3: A profit model

A company's profit, in thousands of dollars, from making and selling

x

hundred units is modeled by

P(x) = -x^3 + 12x^2 - 20x

for

x \geq 0

(a) For what production levels does the company make a profit?

Factor to find where

P(x) = 0

P(x) = -x(x^2 - 12x + 20) = -x(x - 2)(x - 10)

The zeros are

x = 0

x = 2

, and

x = 10

. The sign of

P

can only change at a zero, so test one value in each interval:

Interval	Test value	Sign of $P(x)$
$0 < x < 2$	$P(1) = -9$	Negative (loss)
$2 < x < 10$	$P(5) = 75$	Positive (profit)
$x > 10$	$P(11) = -99$	Negative (loss)

The company makes a profit when

2 < x < 10

, that is, when it makes between 200 and 1,000 units.

(b) What happens to the profit if production keeps increasing?

Use standard form. The leading term is

-x^3

, so

\displaystyle\lim_{x \to \infty} P(x) = -\infty

. In this model, making more and more units eventually leads to bigger and bigger losses.

Polynomial long division

Polynomial long division works like long division with numbers. Dividing 17 by 5 gives a quotient of 3 and a remainder of 2, so

17 = 5 \cdot 3 + 2

. Polynomials work the same way. If the polynomial

f

is divided by the polynomial

g

, then

f(x) = g(x)\,q(x) + r(x)

where

q

is the quotient,

r

is the remainder, and the degree of

r

is less than the degree of

g

. Dividing both sides by

g(x)

gives the form used for rational functions:

\frac{f(x)}{g(x)} = q(x) + \frac{r(x)}{g(x)}

The steps:

Write both polynomials in standard form. If the dividend skips a power, fill it in with a 0 coefficient, like $3x^3 - 5x^2 + 0x + 4$ .

Divide the leading term of what's left by the leading term of the divisor. Write the result in the quotient.

Multiply that new quotient term by the whole divisor.

Subtract the product (change the sign of every term), then bring down the next term.

Repeat until the degree of what's left is less than the degree of the divisor. What's left is the remainder.

Example 4: Dividing by a linear factor

Divide

f(x) = 3x^3 - 5x^2 + 4

g(x) = x - 2

Step 1: Fill in the missing

x

term:

3x^3 - 5x^2 + 0x + 4

Step 2:

3x^3 \div x = 3x^2

. Multiply:

3x^2(x - 2) = 3x^3 - 6x^2

. Subtract:

(3x^3 - 5x^2) - (3x^3 - 6x^2) = x^2

Bring down the

0x

to get

x^2 + 0x

Step 3:

x^2 \div x = x

. Multiply:

x(x - 2) = x^2 - 2x

. Subtract:

(x^2 + 0x) - (x^2 - 2x) = 2x

Bring down the

4

to get

2x + 4

Step 4:

2x \div x = 2

. Multiply:

2(x - 2) = 2x - 4

. Subtract:

(2x + 4) - (2x - 4) = 8

The remainder

8

has degree 0, which is less than the degree of

x - 2

, so we stop. Here's the whole division laid out:

Answer: The quotient is

3x^2 + x + 2

and the remainder is

8

3x^3 - 5x^2 + 4 = (x - 2)(3x^2 + x + 2) + 8

Check: Plug in

x = 2

. The left side is

3(8) - 5(4) + 4 = 8

, and the right side is

0 + 8 = 8

. They match. In general, the remainder after dividing by

x - c

equals

f(c)

Example 5: Dividing by a quadratic

Divide

x^3 + 2x^2 - x + 4

x^2 + 1

Step 1:

x^3 \div x^2 = x

. Multiply:

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

. Subtract:

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

Step 2:

2x^2 \div x^2 = 2

. Multiply:

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

. Subtract:

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

Step 3: The remainder

-2x + 2

has degree 1, which is less than 2, so we stop.

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

The remainder doesn't have to be a number. It only needs a lower degree than the divisor.

Slant asymptotes

When the numerator of a rational function has degree exactly one more than the denominator, there's no horizontal asymptote. Instead, the graph has a slant asymptote (also called an oblique asymptote): a line

y = mx + b

that the graph gets closer and closer to as

x \to \pm\infty

. Long division finds it.

Divide to write

h(x) = \dfrac{f(x)}{g(x)}

h(x) = q(x) + \dfrac{r(x)}{g(x)}

. The remainder has a lower degree than the denominator, so

\dfrac{r(x)}{g(x)} \to 0

x \to \pm\infty

. That leaves

h(x)

very close to

q(x)

, which is linear. The slant asymptote is

y = q(x)

Example 6: Finding a slant asymptote

Find the slant asymptote of

h(x) = \dfrac{x^2 + x - 1}{x - 1}

and describe the graph near it.

Step 1: The numerator has degree 2 and the denominator has degree 1, so there's a slant asymptote.

Step 2: Divide.

x^2 \div x = x

and

x(x - 1) = x^2 - x

. Subtracting leaves

2x - 1

. Next,

2x \div x = 2

and

2(x - 1) = 2x - 2

. Subtracting leaves a remainder of

1

h(x) = x + 2 + \frac{1}{x - 1}

Step 3: As

x \to \pm\infty

\dfrac{1}{x - 1} \to 0

, so the slant asymptote is

y = x + 2

Step 4: The denominator is zero at

x = 1

and the numerator isn't (

1 + 1 - 1 = 1

), so

x = 1

is a vertical asymptote.

The remainder term also tells you which side of the line the graph is on. For large positive

x

\dfrac{1}{x - 1} > 0

, so the graph sits just above

y = x + 2

. For large negative

x

\dfrac{1}{x - 1} < 0

, so it sits just below.

If the numerator's degree is two or more above the denominator's, there's no slant asymptote. The end behavior then follows the quotient, which might be a quadratic or higher.

The binomial theorem and Pascal's triangle

Expanding something like

(x + 2)^5

by multiplying it out five times is slow and easy to mess up. The binomial theorem uses one row of Pascal's triangle to expand

(a + b)^n

directly, including polynomials like

p(x) = (x + c)^n

Building Pascal's triangle

Every row starts and ends with 1. Each entry in between is the sum of the two entries just above it, like

4 + 6 = 10

in red. The rows are numbered starting from row 0.

Row

n

gives the coefficients of

(a + b)^n

. Row 4 is 1, 4, 6, 4, 1, so

(a + b)^4 = 1a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + 1b^4

The pattern

$(a + b)^n$ has $n + 1$ terms.

The powers of $a$ count down from $n$ to 0, and the powers of $b$ count up from 0 to $n$ .

In every term, the exponents add up to $n$ .

The coefficients come from row $n$ of Pascal's triangle.

You may also see the entries of row

n

written as

\binom{n}{0}, \binom{n}{1}, \ldots, \binom{n}{n}

, which gives the general formula below. Reading the coefficients from Pascal's triangle is all AP Precalculus needs.

(a + b)^n = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \cdots + \binom{n}{n}b^n

Example 7: Expanding a power of a binomial

Expand

p(x) = (x + 2)^4

Here

a = x

b = 2

, and

n = 4

. Row 4 is 1, 4, 6, 4, 1.

(x + 2)^4 = 1(x)^4 + 4(x)^3(2) + 6(x)^2(2)^2 + 4(x)(2)^3 + 1(2)^4

= x^4 + 8x^3 + 24x^2 + 32x + 16

Check: Plug in

x = 1

. The left side is

3^4 = 81

, and the right side is

1 + 8 + 24 + 32 + 16 = 81

Example 8: A coefficient and a minus sign

Expand

(2x - 3)^3

Use

a = 2x

b = -3

, and row 3: 1, 3, 3, 1. Keep the parentheses so the 2 and the negative sign get raised to each power.

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

= 8x^3 - 36x^2 + 54x - 27

The signs alternate because odd powers of

-3

are negative.

Example 9: Finding one coefficient (AP-style)

What is the coefficient of

x^2

in the expansion of

(x - 3)^5

Row 5 is 1, 5, 10, 10, 5, 1. The power of

x

counts down from 5, so

x^2

is in the fourth term, where

-3

is raised to the power

5 - 2 = 3

and the coefficient from row 5 is 10:

10(x)^2(-3)^3 = 10 \cdot x^2 \cdot (-27) = -270x^2

Answer: The coefficient is

-270

Common mistakes

Canceling before finding the domain. $\dfrac{(x-2)(x+3)}{(x-2)(x+1)}$ is undefined at $x = 2$ , even though the simplified form isn't. That's exactly why there's a hole.

Calling a canceled factor a vertical asymptote. A factor that cancels completely gives a hole, not an asymptote.

Skipping the zero placeholders. Without the $0x$ in $3x^3 - 5x^2 + 0x + 4$ , the terms in long division line up in the wrong columns.

Subtracting only the first term. When you subtract the product, change the sign of every term in it.

Stopping at the wrong time. Stop exactly when the remainder's degree is less than the divisor's degree.

Dropping a coefficient or sign in the binomial theorem. In $(2x - 3)^3$ , the second term is $3(2x)^2(-3) = -36x^2$ , not $3x^2(-3) = -9x^2$ .

Looking for a slant asymptote when the degrees differ by more than one. A slant asymptote only happens when the numerator's degree is exactly one more than the denominator's.

Back Next