Unit 1 · Topic 1.11

1.11 Equivalent Representations of Polynomial and Rational Expressions

The same polynomial or rational function can be written in several equivalent forms, and each form shows certain features at a glance. You'll pick the useful form, use polynomial long division to split a rational function into a polynomial plus a remainder, and expand powers of binomials with Pascal's triangle.

Key terms

standard form
factored form
polynomial long division
binomial theorem
Pascal's triangle

Which form shows what

Factored form, like 2(x − 1)(x − 3), shows real zeros, and for rational functions it also shows holes, vertical asymptotes and the domain. It can help with the range too.

Standard form, like 2x² − 8x + 6, shows the leading term (so the end behavior) and the y-intercept (the constant term).

For quadratics, vertex form a(x − h)² + k shows the vertex (h, k), which is the global maximum or minimum. All three of these describe the same function: 2x² − 8x + 6 = 2(x − 1)(x − 3) = 2(x − 2)² − 2.

In a context question, choose the form that answers the question asked. If you need when a quantity is zero, factor. If you need its starting value, look at standard form.

Polynomial long division

Dividing polynomial f by polynomial g works like long division with numbers. The result rewrites f as f(x) = g(x)·q(x) + r(x), where q is the quotient and r is the remainder, and the degree of r is less than the degree of g.

Dividing both sides by g gives f(x)/g(x) = q(x) + r(x)/g(x). For large inputs the fraction r(x)/g(x) shrinks toward 0, so the graph of f/g approaches the graph of q. When q is linear, y = q(x) is the slant asymptote.

Steps: divide the leading terms, multiply the result by the whole divisor, subtract, bring down, and repeat until the remainder's degree is lower than the divisor's. Write in every missing power with a 0 coefficient, such as x³ + 0x² − 2x + 4, so the columns line up.

The binomial theorem and Pascal's triangle

To expand (a + b)ⁿ, you don't have to multiply it out. The coefficients come from row n of Pascal's triangle, where each entry is the sum of the two entries above it.

In the expansion, the powers of a count down from n to 0 while the powers of b count up from 0 to n. Each term's exponents add to n.

n	Row of Pascal's triangle
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

Worked examples

Try each one yourself first, then open the solution.

Example 1
Expanding with Pascal's triangle
Expand (x − 2)⁵.
Show the solutionHide the solution
1. Step 1: Use row 5: 1, 5, 10, 10, 5, 1. Here a = x and b = −2.
2. Step 2: Terms: 1·x⁵·(−2)⁰, 5·x⁴·(−2)¹, 10·x³·(−2)², 10·x²·(−2)³, 5·x·(−2)⁴, 1·(−2)⁵.
3. Step 3: Simplify the powers of −2: 1, −2, 4, −8, 16, −32.
4. Step 4: Multiply: x⁵ − 10x⁴ + 40x³ − 80x² + 80x − 32. The signs alternate because b is negative.
Answer: x⁵ − 10x⁴ + 40x³ − 80x² + 80x − 32
Example 2
Long division and a slant asymptote
Divide x³ − 2x² + 4 by x² + 1. Use the result to find the slant asymptote of y = (x³ − 2x² + 4)/(x² + 1).
Show the solutionHide the solution
1. Step 1: Write the dividend with every power: x³ − 2x² + 0x + 4.
2. Step 2: x³ ÷ x² = x. Multiply x(x² + 1) = x³ + x. Subtract: (x³ − 2x² + 0x) − (x³ + x) = −2x² − x. Bring down 4: −2x² − x + 4.
3. Step 3: −2x² ÷ x² = −2. Multiply −2(x² + 1) = −2x² − 2. Subtract: (−2x² − x + 4) − (−2x² − 2) = −x + 6.
4. Step 4: The remainder −x + 6 has degree 1, less than 2, so stop. x³ − 2x² + 4 = (x² + 1)(x − 2) + (−x + 6).
5. Step 5: So y = x − 2 + (−x + 6)/(x² + 1). The fraction approaches 0 at both ends.
Answer: Quotient x − 2, remainder −x + 6; the slant asymptote is y = x − 2.
Example 3
Trap: a coefficient inside the binomial
Expand (2x + 1)⁴.
Show the solutionHide the solution
1. Step 1: Use row 4: 1, 4, 6, 4, 1, with a = 2x and b = 1.
2. Step 2: The whole term 2x gets raised to each power, not just x. A common slip is writing 1·x⁴ for the first term.
3. Step 3: Terms: 1(2x)⁴ = 16x⁴, 4(2x)³(1) = 32x³, 6(2x)²(1) = 24x², 4(2x)(1) = 8x, 1(1)⁴ = 1.
Answer: 16x⁴ + 32x³ + 24x² + 8x + 1

Common mistakes

Raising only the variable to a power in a binomial expansion: (2x)³ is 8x³, not 2x³.
Skipping missing powers in long division, which misaligns the columns. Fill gaps with 0x terms.
Subtracting only the first term during long division. Subtract the whole product, and watch the signs.
Using the quotient of leading terms as the slant asymptote instead of doing the division.

On the exam

No-calculator questions may ask you to rewrite an expression in a form that reveals a feature, such as factoring to find zeros or dividing to find a slant asymptote.
Binomial expansion shows up in multiple choice, sometimes asking for a single coefficient. Find the right row and term rather than expanding everything.

Connected topics

Videos

AP Precalculus – 1.11A Equivalent Expressions and Binomial Theorem
The AlgebrosWatch on YouTube (opens in a new tab)
AP Precalculus – 1.11B Long Division and Slant Asymptotes
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Connecting equivalent forms of rational functions | AP®︎/College Precalculus | Khan Academy
Khan AcademyWatch on YouTube (opens in a new tab)
Representations of Rational Functions [AP Precalculus Topic 1.11]
Michael Porinchak - AP Statistics & AP PrecalculusWatch on YouTube (opens in a new tab)
Polynomial Long Division and Pascals Triangle in Under 3 mins (AP Precalculus Topic 1.11)
Maximum InsightWatch on YouTube (opens in a new tab)
1.11D - Expanding Binomials Using Pascal's Triangle [AP Precalculus]
MrHelpfulNotHurtfulWatch on YouTube (opens in a new tab)