6.2 Dividing Polynomials and the Remainder Theorem

Polynomial Long Division

Polynomial long division works just like the long division you learned with numbers, except now you're dividing expressions with variables. It lets you break a polynomial into a quotient and remainder, which is how you find factors and roots of higher-degree polynomials.

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How It Works

The core idea: repeatedly divide, multiply, and subtract until you can't go any further. Here are the steps:

1. Divide the leading term of the dividend by the leading term of the divisor.
2. Multiply that result by the entire divisor.
3. Subtract the product from the current dividend.
4. Bring down the next term and repeat until the remainder has a lower degree than the divisor.

The result follows this relationship:

$Dividend = (Divisor \times Quotient) + Remainder$

The degree of the quotient equals the degree of the dividend minus the degree of the divisor. So dividing a cubic by a linear term gives you a quadratic quotient.

Example: Divide $x^3 + 2x^2 - 5x + 1$ by $(x - 1)$.

1. $x^3 \div x = x^2$. Multiply: $x^2(x - 1) = x^3 - x^2$. Subtract: $(x^3 + 2x^2) - (x^3 - x^2) = 3x^2$. Bring down $-5x$.

2. $3x^2 \div x = 3x$. Multiply: $3x(x - 1) = 3x^2 - 3x$. Subtract: $(3x^2 - 5x) - (3x^2 - 3x) = -2x$. Bring down $+1$.

3. $-2x \div x = -2$. Multiply: $-2(x - 1) = -2x + 2$. Subtract: $(-2x + 1) - (-2x + 2) = -1$.

Result: $x^3 + 2x^2 - 5x + 1 = (x - 1)(x^2 + 3x - 2) + (-1)$

Divisibility and Factors

If the remainder is 0, the divisor is a factor of the dividend. For example, if dividing $P(x)$ by $(x - 2)$ gives remainder 0, then $P(x) = (x - 2) \cdot Q(x)$ for some quotient $Q(x)$.

This is how you factor higher-degree polynomials: once you know one factor, divide it out, and then try to factor the quotient further using standard techniques.

Remainder Theorem Applications

Evaluating Polynomials

The Remainder Theorem states that when you divide a polynomial $P(x)$ by $(x - a)$, the remainder equals $P(a)$.

This means you can find the remainder without doing long division at all. Just plug in $a$.

Example: For $P(x) = x^3 - 2x^2 + 3x - 4$, find the remainder when dividing by $(x - 2)$:

$P(2) = 2^3 - 2(2^2) + 3(2) - 4 = 8 - 8 + 6 - 4 = 2$

The remainder is 2. No long division needed.

Determining Divisibility

If $P(a) = 0$, then $(x - a)$ divides evenly into $P(x)$. This gives you a quick test for whether a linear expression is a factor.

Example: Is $(x - 1)$ a factor of $P(x) = x^3 - 2x^2 + 3x - 4$?

$P(1) = 1 - 2 + 3 - 4 = -2 \neq 0$

Since the result isn't zero, $(x - 1)$ is not a factor.

Watch the sign carefully: to test $(x + 3)$, you'd evaluate $P(-3)$, because $(x + 3) = (x - (-3))$.

Factoring Polynomials

Factor Theorem

The Factor Theorem is the flip side of the Remainder Theorem: $(x - a)$ is a factor of $P(x)$ if and only if $P(a) = 0$. The "if and only if" means it works both directions.

A polynomial of degree $n$ has exactly $n$ linear factors (counting complex roots and repeated roots). So you can write any polynomial as a product of its linear factors times the leading coefficient.

Example: Factor $P(x) = x^3 - 6x^2 + 11x - 6$.

Test some values:

• $P(1) = 1 - 6 + 11 - 6 = 0$ ✓
• $P(2) = 8 - 24 + 22 - 6 = 0$ ✓
• $P(3) = 27 - 54 + 33 - 6 = 0$ ✓

All three give zero, so: $P(x) = (x - 1)(x - 2)(x - 3)$

In practice, you'd find one root, divide it out, then factor the resulting quadratic.

Using Division to Solve Equations

To solve a polynomial equation like $P(x) = 0$:

1. Find one root $a$ (by testing values or using the Rational Root Theorem).

2. Divide $P(x)$ by $(x - a)$ to get a lower-degree quotient.

3. Factor or solve the quotient to find the remaining roots.

4. Repeat until fully factored.

Remainders and Factors

Rational Root Theorem

When you need candidates to test, the Rational Root Theorem narrows down the possibilities. For a polynomial with integer coefficients:

$a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 = 0$

any rational root $\frac{p}{q}$ (in lowest terms) must satisfy:

• $p$ is a factor of the constant term $a_0$
• $q$ is a factor of the leading coefficient $a_n$

Example: For $2x^3 - 3x^2 + x - 6 = 0$:

• Factors of $a_0 = -6$: $\pm 1, \pm 2, \pm 3, \pm 6$
• Factors of $a_n = 2$: $\pm 1, \pm 2$
• Possible rational roots: $\pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}$

You then test these candidates using the Remainder Theorem (plug them in) or synthetic division. Once you find a root, divide it out and continue factoring.

The typical workflow ties all these tools together: the Rational Root Theorem generates candidates, the Remainder Theorem tests them quickly, and polynomial division breaks the polynomial into smaller factors.