Polynomial Long Division

Concept Explanation

Polynomial long division works exactly like numerical long division. It allows us to break down complex rational fractions into simpler components. The process consists of a repeating cycle of four steps: Divide, Multiply, Subtract, and Bring Down.

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Why We Use It in Calculus

In Calculus, rational expressions can be extremely difficult to analyze in their raw forms. We trigger polynomial division in two primary scenarios:

1. Shortcut for "Factor-and-Cancel" Limits (Calculus I)

When evaluating the limit of a rational function $\lim_{x \to c} \frac{P(x)}{x-c}$, direct substitution often yields the indeterminate form $\frac{0}{0}$. This indicates that $(x-c)$ is a factor of both the numerator and the denominator.

Instead of struggling to factor a high-degree polynomial $P(x)$ by hand, you can perform polynomial long division of $P(x)$ by $(x-c)$. Since $(x-c)$ is a factor, the division will result in a remainder of $0$. You can then cancel out the $(x-c)$ term from the numerator and denominator, allowing you to directly evaluate the limit.

2. Integrating Improper Rational Functions (Calculus II)

An improper rational expression is one where the degree of the numerator is greater than or equal to the degree of the denominator:

We cannot directly apply standard integration techniques (like $u$-substitution) to improper fractions. By performing polynomial division, we rewrite the integrand into a polynomial quotient $Q(x)$ and a proper fraction $\frac{R(x)}{D(x)}$:

• The quotient $Q(x)$ is easily integrated term-by-term using the power rule.
• The proper fraction $\frac{R(x)}{D(x)}$ can then be integrated using simple substitution ($\ln|u|$ forms) or other techniques like partial fractions.

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Example Problems

Example Problem 1: Evaluating a Limit (Factor-and-Cancel Shortcut)

Evaluate the limit using polynomial long division to resolve the indeterminate form:

Solution

Step 1: Check for indeterminate form Substitute $x=2$:

$$\frac{2^3 - 3(2^2) + 5(2) - 6}{2-2} = \frac{8 - 12 + 10 - 6}{0} = \frac{0}{0}$$

This is a $\frac{0}{0}$ indeterminate form, confirming that $(x-2)$ is a factor of the numerator.

Step 2: Perform the division Divide $x^3 - 3x^2 + 5x - 6$ by $x-2$:

1. First term: Divide $x^3$ by $x \implies x^2$.
   • Multiply: $x^2(x - 2) = x^3 - 2x^2$.
   • Subtract: $(x^3 - 3x^2 + 5x - 6) - (x^3 - 2x^2) = -x^2 + 5x - 6$.
2. Second term: Divide $-x^2$ by $x \implies -x$.
   • Multiply: $-x(x - 2) = -x^2 + 2x$.
   • Subtract: $(-x^2 + 5x - 6) - (-x^2 + 2x) = 3x - 6$.
3. Third term: Divide $3x$ by $x \implies 3$.
   • Multiply: $3(x - 2) = 3x - 6$.
   • Subtract: $(3x - 6) - (3x - 6) = 0$.

The division has a remainder of $0$, giving:

$$\frac{x^3 - 3x^2 + 5x - 6}{x - 2} = x^2 - x + 3 \quad (\text{for } x \neq 2)$$

Step 3: Evaluate the limit Now, substitute $x=2$ into the simplified quotient:

$$\lim_{x \to 2} (x^2 - x + 3) = 2^2 - 2 + 3 = 4 - 2 + 3 = 5$$

The limit is $5$.

Example Problem 2: Integration Preparation

Rewrite the integrand using polynomial long division, then evaluate the indefinite integral:

Solution

Step 1: Check the degrees The degree of the numerator is $3$, and the degree of the denominator is $1$. Since $3 \geq 1$, this is an improper rational function and we must divide.

Step 2: Perform the division Divide $2x^3 - 3x^2 + 4x + 5$ by $x-2$:

1. First term: Divide $2x^3$ by $x \implies 2x^2$.
   • Multiply: $2x^2(x - 2) = 2x^3 - 4x^2$.
   • Subtract: $(2x^3 - 3x^2 + 4x + 5) - (2x^3 - 4x^2) = x^2 + 4x + 5$.
2. Second term: Divide $x^2$ by $x \implies x$.
   • Multiply: $x(x - 2) = x^2 - 2x$.
   • Subtract: $(x^2 + 4x + 5) - (x^2 - 2x) = 6x + 5$.
3. Third term: Divide $6x$ by $x \implies 6$.
   • Multiply: $6(x - 2) = 6x - 12$.
   • Subtract: $(6x + 5) - (6x - 12) = 17$.

The quotient is $2x^2 + x + 6$ and the remainder is $17$.

Step 3: Rewrite the integral

$$\int \left(2x^2 + x + 6 + \frac{17}{x - 2}\right) \, dx$$

Step 4: Integrate term-by-term

$$\int (2x^2) \, dx + \int (x) \, dx + \int (6) \, dx + \int \left(\frac{17}{x - 2}\right) \, dx$$

$$= \frac{2}{3}x^3 + \frac{1}{2}x^2 + 6x + 17\ln|x - 2| + C$$

The value of the integral is $\frac{2}{3}x^3 + \frac{1}{2}x^2 + 6x + 17\ln|x - 2| + C$.

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Connections & References

• Parent Concepts: Calculus Foundations
• Related Concepts: Factoring Second Order Polynomials
• Course Links: Calc-1 Discussion 01