Topic 1.11A · CED: Equivalent Representations of Polynomial and Rational Expressions

Equivalent Representations & Binomial Theorem

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01
Factored form is almost always more useful for analysis — expand only when a problem specifically asks for standard/general form.

Practice 22 problems Next: Chapter 1.11B Polynomial Long Division and Slant Asymptotes

Factored Form vs. Standard/General Form

Every polynomial and rational function can be written in more than one equivalent way. Factored form directly shows you zeros, holes, and asymptotes. Standard/general form (fully multiplied out) is what you get after expanding — useful for some tasks, but it hides the very features factored form reveals instantly.

Polynomial Function

CONCEPT

Factored Form

f(x) = (x + 6)^2 (x - 1)^3

CONCEPT

Standard Form (fully expanded)

f(x) = x^5 + 9x^4 + 3x^3 - 73x^2 + 96x - 36

Rational Function

CONCEPT

Factored Form

f(x) = \frac{(x + 1)(x - 6)}{(x + 1)(x + 5)}

CONCEPT

General Form (numerator and denominator expanded)

f(x) = \frac{x^2 - 5x - 6}{x^2 + 6x + 5}

COMMON MISTAKE

Writing the CANCELED expression $\dfrac{x - 6}{x + 5}$ as the general form.

Canceling $(x+1)$ throws away the hole at $x = -1$ . The canceled expression is defined at $x = -1$ ; the original is not — so the two are not the same function, and one is not the "general form" of the other.

General form means EXPANDED, not SIMPLIFIED. Multiply out; do not cancel.

KEY RULE

Factored form is almost always more useful for analysis — expand only when a problem specifically asks for standard/general form.

Quick check

$f(x) = \frac{x(x - 4)}{(x - 4)(2x + 1)}$ . Is $\frac{x}{2x + 1}$ the same function as $f$ ?

Reading Everything from a Polynomial's Factored Form

Worked example

For $g(x) = 3x^{3} - 18x^{2} + 27x$ , find: factored form, degree, end behavior, y-intercept, zero(s), and where $g(x) \ge 0$ .

g(x) = 3x^3 - 18x^2 + 27x

01
Factor out the GCF, then factor what's left:
$g(x) = 3x(x^2 - 6x + 9) = 3x(x - 3)^2$
02
Degree = 3 (add the exponents: 1 from $x$ , 2 from $(x-3)^{2}$ ).
03
End behavior: odd degree, positive leading coefficient (3) → down on the left, up on the right.
04
y-intercept: $g(0) = 0$ (plug in $x=0$ , or just notice $x$ is a factor).
05
Zeros: $x=0$ (multiplicity 1, crosses) and $x=3$ (multiplicity 2, bounces).
06
Where is $g(x) \ge 0$ ? Since $x=3$ is a bounce (sign doesn't flip there) and $x=0$ is a simple crossing, testing intervals shows $g(x) \ge 0$ on $[0, \infty)$ .

Crosses at $x=0$ , bounces at $x=3$ — confirming $g(x)$ stays non-negative from $x=0$ onward.

Reading Everything from a Rational Function's Factored Form

Worked example

For $h(x) = 4x(x-6) / [(x+2)(x-3)]$ , find: domain, zero(s), hole(s), vertical asymptote(s), horizontal asymptote, and y-intercept.

h(x) = \frac{4x(x - 6)}{(x + 2)(x - 3)}

01
No common factors between numerator and denominator → no holes.
02
Domain: all reals except $x=-2$ and $x=3$ (denominator zeros).
03
Zeros: $x=0$ and $x=6$ (numerator zeros, both safely in the domain).
04
Vertical asymptotes: $x=-2$ and $x=3$ .
05
Horizontal asymptote: both numerator and denominator are degree 2 (once expanded) with leading coefficients 4 and 1 → $y = 4$ .
06
y-intercept: $h(0) = 4(0)(-6) / [(2)(-3)] = 0/(-6) = 0$ .

Every feature read straight from the factored form — no expanding required.

COMMON MISTAKE

Don't expand a factored expression before checking for common factors — you'll lose the ability to spot holes easily.

Degree and leading coefficient for a horizontal asymptote can be found from factored form too — just add up exponents and multiply leading coefficients, no need to fully expand.

The Binomial Theorem: Building Intuition with Pascal's Triangle

Expanding a binomial by hand gets tedious fast. Let's spot the pattern by expanding the same binomial to increasing powers:

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

(x + 2)^3 = x^3 + 6x^2 + 12x + 8

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

CONCEPT

Spot the Pattern: Pascal's Triangle

The coefficients (1,2,1), then (1,3,3,1), then (1,4,6,4,1) are rows of Pascal's Triangle — each entry is the sum of the two entries above it.

This pattern holds for ANY binomial raised to ANY power, not just $(x+2)$ .

\begin{array}{rccccccc} n=0 & & & & 1 \\ n=1 & & & 1 & & 1 \\ n=2 & & 1 & & 2 & & 1 \\ n=3 & 1 & & 3 & & 3 & & 1 \\ n=4 & 1 & 4 & & 6 & & 4 & 1 \\ n=5 & \boxed{1} & \boxed{5} & \boxed{10} & \boxed{10} & \boxed{5} & \boxed{1} \\ n=6 & 1 & 6 & 15 & 20 & 15 & 6 & 1 \end{array}

Each row of Pascal's Triangle gives the coefficients for expanding a binomial to that power — every number is the sum of the two above it.

The Formal Binomial Theorem

CONCEPT

Binomial Theorem

The expansion of a binomial is:

(a + b)^n = C_1 a^n b^0 + C_2 a^{n-1} b^1 + C_3 a^{n-2} b^2 + \cdots + C_n a^0 b^n

where $C$ is the coefficient given by Pascal's Triangle (equivalently, C = ₙCₖ, the combinations formula).

Step-by-Step: Full Expansion

Worked example

Expand $(x - 2)^{5}$ .

(x - 2)^5

01
Row 5 of Pascal's Triangle gives the coefficients: 1, 5, 10, 10, 5, 1.
02
Write each term with decreasing powers of $x$ and increasing powers of −2, keeping exponents summing to 5 in every term. Here's exactly how each coefficient and each term is built:
$\begin{array}{r|cccccc} \text{row 5} & 1 & 5 & 10 & 10 & 5 & 1 \\ \text{parts} & x^5 & x^4(-2)^1 & x^3(-2)^2 & x^2(-2)^3 & x^1(-2)^4 & (-2)^5 \\ \hline \text{result} & x^5 & -10x^4 & 40x^3 & -80x^2 & 80x & -32 \end{array}$
Each Pascal's Triangle number becomes a coefficient; each term pairs a power of $x$ with a power of −2.
03
Combine and simplify:
$x^5 - 10x^4 + 40x^3 - 80x^2 + 80x - 32$

Step-by-Step: Finding One Specific Term

Worked example

Find the third term in the expansion of $(3x + 2)^{6}$ — without expanding the whole thing.

(3x + 2)^6, \;\text{ find the 3rd term}

01
The third term corresponds to $k=2$ (the first term is $k=0$ ). Use the general term formula C(n,k)·a^(n−k)·b^k with $a=3x$ , $b=2$ , $n=6$ :
$C(6,\, 2)\,(3x)^4 (2)^2$
02
Compute $C(6, 2) = 15$ , (3x)⁴=81x⁴, 2²=4:
$15 \cdot 81x^4 \cdot 4 = 4860x^4$
03
The third term is $4860x^{4}$ .

COMMON MISTAKE

The exponents in each term must always add up to $n$ — double-check this in every term you write.

When finding a specific term, the term number is $k+1$ , not $k$ — the FIRST term corresponds to $k=0$ , so the THIRD term uses $k=2$ .

Don't forget to raise the entire second quantity (including any coefficient, like the '2' in $(3x+2)$ ) to its power — a common shortcut error is only using the variable part.

Quick check

In the expansion of $(x + 2)^{9}$ with decreasing powers of $x$ , which $k$ gives the third term, and what is it?

Practice Problems

Worked example

Problem 1. Expand $(x + 3)^{4}$ using the Binomial Theorem.

01
Row 4 of Pascal's Triangle: 1, 4, 6, 4, 1.
02
$(x+3)^{4} = x^{4} + 4x^{3}(3) + 6x^{2}(9) + 4x(27) + 81 = x^{4}+12x^{3}+54x^{2}+108x+81$ .

Worked example

Problem 2. Find the fourth term in the expansion of $(x - 5)^{7}$ .

01
Fourth term → $k=3$ . $C(7, 3) = 35$ . Term = $35 \cdot x^{4} \cdot (-5)^{3} = 35 \cdot x^{4} \cdot (-125) = -4375x^{4}$ .

Common slips

Writing the canceled expression $\dfrac{x - 6}{x + 5}$ as the general form.

Canceling $(x+1)$ throws away the hole at $x = -1$ . The canceled expression is defined at $x = -1$ ; the original is not — so the two are not the same function, and one is not the "general form" of the other.

General form means expanded, not simplified. Multiply out; do not cancel.
Don't expand a factored expression before checking for common factors — you'll lose the ability to spot holes easily.

Degree and leading coefficient for a horizontal asymptote can be found from factored form too — just add up exponents and multiply leading coefficients, no need to fully expand.
The exponents in each term must always add up to $n$ — double-check this in every term you write.

When finding a specific term, the term number is $k+1$ , not $k$ — the first term corresponds to $k=0$ , so the third term uses $k=2$ .

Don't forget to raise the entire second quantity (including any coefficient, like the '2' in $(3x+2)$ ) to its power — a common shortcut error is only using the variable part.

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7 cards · Equivalent forms and division

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Recap card

4 lines to re-read the night before.

01
Factored form reveals zeros, holes, and asymptotes directly; standard/general form hides them behind expanded terms.
02
Degree, end behavior, and horizontal asymptotes can all be read from factored form without fully expanding.
03
The Binomial Theorem expands $(a+b)^{n}$ using coefficients from Pascal's Triangle.
04
A single term can be found directly with C(n,k)·a^(n−k)·b^k — no need to expand the whole binomial.

Practice 22 problems Next: Chapter 1.11B Polynomial Long Division and Slant Asymptotes

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Equivalent Representations & Binomial Theorem

Factored Form vs. Standard/General Form

Polynomial Function

Rational Function

Reading Everything from a Polynomial's Factored Form

Reading Everything from a Rational Function's Factored Form

The Binomial Theorem: Building Intuition with Pascal's Triangle

The Formal Binomial Theorem

Step-by-Step: Full Expansion

Step-by-Step: Finding One Specific Term

Practice Problems

Recap card

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