3.3 Polynomial Division and the Remainder Theorem

3.3 Polynomial Division and the Remainder Theorem

Dividing polynomials is analogous to long division of integers. When an integer division is not exact, the result is expressed as a quotient plus a fractional remainder (e.g., $17 \div 5 = 3 + \frac{2}{5}$). Polynomial division behaves identically: when dividing dividend $P(x)$ by divisor $D(x)$, the algebraic result can always be written in the Division Algorithm format.

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The Division Algorithm for Polynomials

For any polynomial dividend $P(x)$ and non-zero polynomial divisor $D(x)$, there exist unique polynomials $Q(x)$ (the quotient) and $R(x)$ (the remainder) such that:

$$P(x) = D(x) \cdot Q(x) + R(x)$$

Dividing both sides by the divisor $D(x)$ yields the rational form required on Regents exams:

$$\frac{P(x)}{D(x)} = Q(x) + \frac{R(x)}{D(x)}$$

Where either $R(x) = 0$ or the degree of $R(x)$ is strictly less than the degree of $D(x)$:

$$\deg(R) < \deg(D)$$

If the divisor is linear ($D(x) = x - c$, degree 1), the remainder must have degree 0—meaning $R(x)$ is simply a constant number $R$.

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Polynomial Long Division Step-by-Step

Polynomial long division works for divisors of any degree (linear, quadratic, or higher). The division algorithm follows four recurring steps: Divide $\to$ Multiply $\to$ Subtract $\to$ Bring Down.

Worked Example: Long Division

Divide $2x^3 - 3x^2 + 5x - 4$ by $x - 2$.

               2x² +  x  + 7     <── Quotient Q(x)
         ┌─────────────────────────
  x - 2  │  2x³ - 3x² + 5x - 4
          -(2x³ - 4x²)           <── Subtract 2x²(x - 2)
         ─────────────────
                   x² + 5x
                 -(x² - 2x)      <── Subtract x(x - 2)
                 ─────────
                       7x - 4
                     -(7x - 14)  <── Subtract 7(x - 2)
                     ─────────
                            10   <── Remainder R

1. First Cycle:
   • Divide: Divide leading terms: $\frac{2x^3}{x} = 2x^2$.
   • Multiply: Multiply $2x^2$ by divisor: $2x^2(x - 2) = 2x^3 - 4x^2$.
   • Subtract: $(2x^3 - 3x^2) - (2x^3 - 4x^2) = x^2$.
   • Bring Down: Bring down $+5x$ to form the new dividend $x^2 + 5x$.
2. Second Cycle:
   • Divide: $\frac{x^2}{x} = x$.
   • Multiply: $x(x - 2) = x^2 - 2x$.
   • Subtract: $(x^2 + 5x) - (x^2 - 2x) = 7x$.
   • Bring Down: Bring down $-4$ to form $7x - 4$.
3. Third Cycle:
   • Divide: $\frac{7x}{x} = 7$.
   • Multiply: $7(x - 2) = 7x - 14$.
   • Subtract: $(7x - 4) - (7x - 14) = 10$.
4. Final Formulation: $$\frac{2x^3 - 3x^2 + 5x - 4}{x - 2} = 2x^2 + x + 7 + \frac{10}{x - 2}$$

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Synthetic Division: The Efficient Linear Shortcut

Synthetic division is a streamlined algorithm that eliminates variables and focuses solely on coefficients. However, it can be applied only when the divisor is a first-degree linear polynomial of the form $x - c$.

Synthetic Division Mechanics for $(x - c)$

To divide $P(x)$ by $x - c$:

1. Place the root $c$ in the corner box (note the sign reversal: if dividing by $x - 2$, $c = 2$; if dividing by $x + 3$, $c = -3$).
2. Write down the coefficients of $P(x)$ in order of descending powers, including zeros for missing terms.
3. Drop the first leading coefficient directly down below the line.
4. Multiply that number by $c$, write the product under the next coefficient, and add down the column.
5. Repeat until all columns are complete. The final number is the remainder; the preceding numbers are the coefficients of the quotient (of degree $n - 1$).

  2 │   2   -3    5   -4
    │        4    2   14
    └────────────────────
        2    1    7   10  ───> Quotient: 2x² + x + 7, Remainder: 10

Handling Divisors with Leading Coefficients: $ax - b$

When the divisor is of the form $ax - b$ with $a \neq 1$ (such as $2x - 3$ or $3x + 1$), students frequently make severe errors. Synthetic division assumes a monic divisor $(x - c)$.

$$\frac{P(x)}{ax - b} = \frac{P(x)}{a\left(x - \frac{b}{a}\right)} = \frac{1}{a} \cdot \left[\frac{P(x)}{x - \frac{b}{a}}\right]$$

Worked Example: Dividing by $2x - 3$

Divide $(6x^3 + 5x^2 - 17x + 10)$ by $(2x - 3)$:

1. Find the root: $2x - 3 = 0 \implies x = \frac{3}{2}$. Set $c = \frac{3}{2}$.
2. Set up synthetic division:

 3/2 │   6    5   -17   10
     │        9    21    6
     └─────────────────────
         6   14     4   16

1. Adjust the quotient coefficients: Divide the bottom row (except the remainder) by $a = 2$: $$\frac{6}{2} = 3, \quad \frac{14}{2} = 7, \quad \frac{4}{2} = 2$$
2. Assemble the result:
   • Adjusted Quotient: $Q(x) = 3x^2 + 7x + 2$
   • Remainder: $R = 16$
3. Write in standard rational form: $$\frac{6x^3 + 5x^2 - 17x + 10}{2x - 3} = 3x^2 + 7x + 2 + \frac{16}{2x - 3}$$

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The Remainder Theorem

The Remainder Theorem provides an elegant bridge between polynomial division and functional evaluation.

The Remainder Theorem: If a polynomial $P(x)$ is divided by a linear binomial $(x - c)$, then the resulting constant remainder is equal to $P(c)$.

Mathematical Proof

From the Division Algorithm:

$$P(x) = (x - c) \cdot Q(x) + R$$

Since the divisor $(x - c)$ is degree 1, the remainder $R$ is a degree 0 constant. Evaluate this identity at the specific value $x = c$:

$$P(c) = (c - c) \cdot Q(c) + R$$ $$P(c) = 0 \cdot Q(c) + R$$ $$P(c) = R \quad \blacksquare$$

Application: Synthetic Substitution

Because evaluating high-degree polynomials by direct substitution can be cumbersome and error-prone, you can use synthetic division to find $P(c)$ quickly. The final remainder of synthetic division is the exact value of $P(c)$.

Example:

To evaluate $P(x) = 3x^3 - 5x^2 + 4x - 7$ at $x = 2$:

  2 │   3   -5    4   -7
    │        6    2   12
    └────────────────────
        3    1    6    5  ───> P(2) = 5

Check via direct substitution: $P(2) = 3(8) - 5(4) + 4(2) - 7 = 24 - 20 + 8 - 7 = 5$. Both methods yield exactly 5.

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The Factor Theorem

The Factor Theorem is an immediate, powerful corollary of the Remainder Theorem:

The Factor Theorem: A linear binomial $(x - c)$ is a factor of a polynomial $P(x)$ if and only if $P(c) = 0$.

This theorem establishes complete equivalence among four fundamental algebraic concepts:

$$\text{Root of Equation } P(x) = 0 \iff \text{Zero of Function } P(c) = 0 \iff x\text{-Intercept at } (c, 0) \iff \text{Linear Factor } (x - c)$$

Application: Finding an Unknown Parameter $k$

A signature problem type on the NY Regents examination requires students to find an unknown coefficient $k$ that makes a given binomial an exact factor.

Model Problem: Solve for Parameter $k$

For what value of $k$ will $(x + 2)$ be a factor of $f(x) = x^3 + kx^2 - 4x + 12$?

• Step 1: Identify the divisor and root: Divisor is $(x + 2) = (x - (-2))$, so $c = -2$.
• Step 2: Apply the Factor Theorem: For $(x + 2)$ to be a factor, the remainder must equal zero, meaning $f(-2) = 0$.
• Step 3: Substitute $x = -2$ into $f(x)$ and set equal to 0: $$(-2)^3 + k(-2)^2 - 4(-2) + 12 = 0$$
• Step 4: Simplify each term: $$-8 + 4k + 8 + 12 = 0$$
• Step 5: Combine like terms: $$(-8 + 8) + 4k + 12 = 0 \implies 4k + 12 = 0$$
• Step 6: Solve for $k$: $$4k = -12 \implies k = -3$$
• Verification: Substitute $k = -3$ back into $f(x)$: $f(x) = x^3 - 3x^2 - 4x + 12$. Factor by grouping: $x^2(x - 3) - 4(x - 3) = (x^2 - 4)(x - 3) = (x - 2)(x + 2)(x - 3)$. The factor $(x + 2)$ is confirmed, proving $k = -3$ is correct.

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Comparison of Division and Root-Finding Methods