Factor and Cancel

Core Definition

The Factor and Cancel strategy is an algebraic limit evaluation method used when direct substitution into a rational function $\frac{P(x)}{Q(x)}$ yields the indeterminate form $\frac{0}{0}$ as $x \to a$ .

The Target Factor Principle: Because $x \to a$ , the factor causing division by zero is always $(x - a)$ . Factoring or dividing out $(x - a)$ from both the numerator and denominator removes the removable discontinuity, enabling direct substitution:

\lim_{x \to a} \frac{P(x)}{Q(x)} = \lim_{x \to a} \frac{(x - a) \cdot \tilde{P}(x)}{(x - a) \cdot \tilde{Q}(x)} = \lim_{x \to a} \frac{\tilde{P}(x)}{\tilde{Q}(x)} = \frac{\tilde{P}(a)}{\tilde{Q}(a)}

Concept Explanation

When direct substitution of $x = a$ produces $\frac{0}{0}$ , the Polynomial Remainder Theorem guarantees that $x = a$ is a root of both the numerator $P(x)$ and denominator $Q(x)$ . Therefore, $(x - a)$ is guaranteed to be an exact factor of both expressions.

Because a limit analyzes behavior near $a$ rather than at $a$ ( $x \neq a$ ), the term $(x - a) \neq 0$ . This allows us to divide $(x - a)$ out cleanly without violating rules of arithmetic.

       Direct Substitution
        x = a  --->  0 / 0  (Indeterminate Hole)
               |
     +---------+---------+
     | Target Factor is  |
     |   always (x - a)  |
     +---------+---------+
               |
     +---------+-------------------------+
     |                                   |
Method 1: Factoring              Method 2: Long Division
Trial & error / AC method /      Divide polynomial by (x - a) via
difference of squares            <a class="text-accent underline font-semibold" href="/garden/polynomial-long-division">Polynomial Long Division</a>
     |                                   |
     +-----------------+-----------------+
                       |
               Cancel (x - a)
                       |
             Re-substitute x = a

Two Methods to Extract $(x - a)$

Method 1: Traditional Polynomial Factoring

Use standard algebraic factoring patterns (factoring techniques, difference of squares, greatest common factor, or the AC method for quadratics) to rewrite $P(x)$ and $Q(x)$ as products of binomials.

Method 2: The Polynomial Long Division Shortcut

When factoring is difficult—such as with higher-degree cubics, non-monic quadratics ( $ax^2+bx+c$ ), or polynomials in the numerator or denominator where factoring isn't obvious—you can use Polynomial Long Division (or synthetic division) to divide $P(x)$ or $Q(x)$ directly by $(x - a)$ .

Because $x = a$ is guaranteed to be a root, the division will always produce a remainder of $0$ . The resulting quotient is the simplified polynomial, allowing you to bypass factoring guesses completely.

Example Problems

Example Problem 1: Factoring vs. Long Division Comparison (Quadratic)

Evaluate the limit:

\lim_{x \to 2} \frac{x^2 + 3x - 10}{x - 2}

Solution

Step 1: Check direct substitution

\frac{2^2 + 3(2) - 10}{2 - 2} = \frac{4 + 6 - 10}{0} = \frac{0}{0} \quad (\text{Indeterminate})

Since

x \to 2

, the target factor causing

\frac{0}{0}

is $(x - 2)$ .

Method A — Traditional Factoring:
Find factors of $-10$ that sum to $+3$ : numbers are $+5$ and $-2$ .

\lim_{x \to 2} \frac{(x - 2)(x + 5)}{x - 2} = \lim_{x \to 2} (x + 5) = 2 + 5 = 7

Method B — Polynomial Long Division:
Divide the numerator $(x^2 + 3x - 10)$ directly by the known factor $(x - 2)$ :

$x^2 \div x = x$
Subtract $x(x - 2) = x^2 - 2x \implies 5x - 10$
$5x \div x = 5$
Subtract $5(x - 2) = 5x - 10 \implies \text{Remainder } 0$

Quotient is $(x + 5)$ . Substitute $x = 2$ : $2 + 5 = 7$ .

Answer: $7$

Example Problem 2: Higher Degree Cubic Polynomial in Numerator

Evaluate the limit:

\lim_{x \to 3} \frac{x^3 - 2x^2 - 4x + 3}{x - 3}

Solution

Step 1: Check direct substitution

\frac{3^3 - 2(3^2) - 4(3) + 3}{3 - 3} = \frac{27 - 18 - 12 + 3}{0} = \frac{0}{0}

The target factor to cancel is $(x - 3)$ .

Step 2: Use Polynomial Long Division
Divide $x^3 - 2x^2 - 4x + 3$ by $(x - 3)$ :

$x^3 \div x = x^2$ . Multiply: $x^2(x - 3) = x^3 - 3x^2$ . Subtract $\implies x^2 - 4x + 3$ .
$x^2 \div x = x$ . Multiply: $x(x - 3) = x^2 - 3x$ . Subtract $\implies -x + 3$ .
$-x \div x = -1$ . Multiply: $-1(x - 3) = -x + 3$ . Subtract $\implies \text{Remainder } 0$ .

The quotient is $x^2 + x - 1$ .

Step 3: Evaluate simplified limit

\lim_{x \to 3} (x^2 + x - 1) = 3^2 + 3 - 1 = 9 + 3 - 1 = 11

Answer: $11$

Example Problem 3: Higher Degree Polynomial in Denominator

Evaluate the limit:

\lim_{x \to 2} \frac{x^2 + x - 6}{x^3 - x^2 - x - 2}

Solution

Step 1: Check direct substitution

Numerator: $2^2 + 2 - 6 = 0$
Denominator: $2^3 - 2^2 - 2 - 2 = 8 - 4 - 2 - 2 = 0$ This yields $\frac{0}{0}$ . Since $x \to 2$ , both polynomials share the factor $(x - 2)$ .

Step 2: Factor numerator and divide denominator

Numerator factors easily: $x^2 + x - 6 = (x - 2)(x + 3)$ .
Denominator is a cubic: divide $x^3 - x^2 - x - 2$ by $(x - 2)$ via polynomial long division:
1. $x^3 \div x = x^2 \implies x^3 - x^2 - (x^3 - 2x^2) = x^2 - x - 2$ .
2. $x^2 \div x = x \implies x^2 - x - (x^2 - 2x) = x - 2$ .
3. $x \div x = 1 \implies (x - 2) - (x - 2) = 0$ (Remainder $0$ ).
Denominator quotient is $(x^2 + x + 1)$ .

Step 3: Cancel $(x - 2)$ and evaluate

\lim_{x \to 2} \frac{(x - 2)(x + 3)}{(x - 2)(x^2 + x + 1)} = \lim_{x \to 2} \frac{x + 3}{x^2 + x + 1} = \frac{2 + 3}{2^2 + 2 + 1} = \frac{5}{7}

Answer: $\frac{5}{7}$

Connections & References

Parent Concepts: Limits, Calculus Foundations
Related Concepts: Polynomial Long Division, Factoring Second Order Polynomials, Removable Discontinuity, Radical Conjugate Rationalization
Course Links: Calc-1 Session 03

Tony's Intertwingled Memex • Reference Manual

Concept Explanation

Two Methods to Extract (x−a)(x - a)(x−a)

Method 1: Traditional Polynomial Factoring

Method 2: The Polynomial Long Division Shortcut

Example Problems

Example Problem 1: Factoring vs. Long Division Comparison (Quadratic)

Example Problem 2: Higher Degree Cubic Polynomial in Numerator

Example Problem 3: Higher Degree Polynomial in Denominator

Connections & References

Two Methods to Extract $(x - a)$