Math Core

Division and evaluation look like different jobs, but for polynomials they're secretly the same. That connection gives you a fast way to test whether $x - c$ is a factor of a polynomial, which is the key step in factoring cubics and quartics.

The remainder theorem

Divide $P(x) = x^3 - 4x^2 + x + 7$ by $x - 3$ using synthetic division:

\begin{array}{r|rrrr} 3 & 1 & -4 & 1 & 7 \\ & & 3 & -3 & -6 \\ \hline & 1 & -1 & -2 & 1 \end{array}

The remainder is $1$ . Now evaluate $P(3)$ directly:

P(3) = 27 - 36 + 3 + 7 = 1.

The same number. That's not a coincidence.

Remainder theorem

When a polynomial $P(x)$ is divided by $x - c$ , the remainder is $P(c)$ .

Why it works. By the division algorithm, $P(x) = (x - c)\,Q(x) + R$ , where $R$ is a constant because the divisor is linear. Substitute $x = c$ :

P(c) = (c - c)\,Q(c) + R = 0 \cdot Q(c) + R = R.

The factor $x - c$ becomes $0$ and wipes out the quotient, leaving only the remainder.

The theorem works in both directions, and both are useful:

To find a remainder, you can just plug in $c$ . That's often faster than dividing.
To evaluate a polynomial at a messy input, synthetic division is often faster than plugging in, especially for high degrees. Used this way it's called synthetic substitution.

Worked example: Two ways to get a remainder

Find the remainder when $P(x) = 2x^4 + 3x^3 - x + 6$ is divided by $x + 2$ .

The divisor is $x - (-2)$ , so the remainder is $P(-2)$ :

P(-2) = 2(16) + 3(-8) - (-2) + 6 = 32 - 24 + 2 + 6 = 16.

Synthetic division agrees:

\begin{array}{r|rrrrr} -2 & 2 & 3 & 0 & -1 & 6 \\ & & -4 & 2 & -4 & 10 \\ \hline & 2 & -1 & 2 & -5 & 16 \end{array}

The remainder is $16$ .

Common mistake

The divisor $x + 2$ means you evaluate at $x = -2$ , not $x = 2$ . Rewrite the divisor as $x - c$ first so you read off $c$ with the right sign.

The factor theorem

A remainder of $0$ means the division comes out even, so the divisor is a factor. Combining that with the remainder theorem gives one of the most useful facts in the unit.

Factor theorem

$x - c$ is a factor of $P(x)$ if and only if $P(c) = 0$ .

In other words, the zeros of a polynomial and its linear factors match up: $c$ is a zero exactly when $x - c$ is a factor. If you know that $P(5) = 0$ , you know $P(x) = (x - 5)\,Q(x)$ for some polynomial $Q$ of one lower degree. And if you know a factor, you know a zero.

Worked example: Testing and using a factor

Is $x + 2$ a factor of $P(x) = x^3 + 3x^2 - 4$ ? If so, factor $P(x)$ completely.

Test $c = -2$ :

P(-2) = -8 + 12 - 4 = 0.

So $x + 2$ is a factor. Divide to find the other factor:

\begin{array}{r|rrrr} -2 & 1 & 3 & 0 & -4 \\ & & -2 & -2 & 4 \\ \hline & 1 & 1 & -2 & 0 \end{array}

The quotient is $x^2 + x - 2 = (x + 2)(x - 1)$ . So

x^3 + 3x^2 - 4 = (x + 2)(x + 2)(x - 1) = (x + 2)^2 (x - 1).

The zeros are $-2$ and $1$ . Notice the factor $x + 2$ appears twice.

Finding a missing coefficient

If you're told that some $x - c$ is a factor, then $P(c) = 0$ gives you an equation. That lets you solve for an unknown coefficient.

Worked example: Solving for k

Find $k$ so that $x - 2$ is a factor of $P(x) = x^3 + kx^2 - 5x + 6$ .

By the factor theorem, you need $P(2) = 0$ :

\begin{aligned} 2^3 + k(2^2) - 5(2) + 6 &= 0 \\ 8 + 4k - 10 + 6 &= 0 \\ 4k + 4 &= 0 \\ k &= -1. \end{aligned}

Check: with $k = -1$ , $P(2) = 8 - 4 - 10 + 6 = 0$ . ✓

The same idea works with remainders. "Dividing by $x - 3$ leaves remainder $7$ " just means $P(3) = 7$ . Two such facts give two equations, which you can solve as a system.

Factoring completely from one known factor

Once you have one linear factor of a cubic, dividing it out leaves a quadratic, and you already know how to factor or solve quadratics. This is the standard way to break down a cubic.

Worked example: From one factor to all the zeros

Given that $x - 3$ is a factor of $P(x) = 2x^3 - 3x^2 - 11x + 6$ , factor $P(x)$ completely and list its zeros.

Divide by $x - 3$ :

\begin{array}{r|rrrr} 3 & 2 & -3 & -11 & 6 \\ & & 6 & 9 & -6 \\ \hline & 2 & 3 & -2 & 0 \end{array}

The remainder is $0$ , as promised. Factor the quotient $2x^2 + 3x - 2$ by looking for two numbers with product $2 \cdot (-2) = -4$ and sum $3$ : those are $4$ and $-1$ .

2x^2 + 3x - 2 = 2x^2 + 4x - x - 2 = 2x(x + 2) - (x + 2) = (2x - 1)(x + 2).

2x^3 - 3x^2 - 11x + 6 = (x - 3)(2x - 1)(x + 2),

and the zeros are $3$ , $\tfrac{1}{2}$ and $-2$ . (The factor $2x - 1$ gives the zero $x = \tfrac{1}{2}$ .)

Tip

The bottom row of synthetic division is a built-in check. If you were told $x - c$ is a factor and the last number isn't $0$ , you've made an arithmetic slip, so find it before moving on.

Practice

Practice 1

Find the remainder when $x^3 - 4x^2 + 2x + 5$ is divided by $x - 2$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

Find the remainder when $2x^4 - x^3 + 3x - 10$ is divided by $x + 1$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

Which of these is a factor of $x^3 - 7x + 6$ ?

Practice 4

Find $k$ so that $x + 2$ is a factor of $x^3 - 2x^2 + kx + 12$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

Given that $x - 4$ is a factor of $x^3 - 3x^2 - 6x + 8$ , find all the zeros of the polynomial.

Separate answers with commas, e.g. 2, -5

Practice 6

Use synthetic substitution to find $P(5)$ for $P(x) = x^4 - 6x^3 + 8x^2 - 5x + 3$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

Given that $2$ is a zero of $P(x) = 2x^3 + x^2 - 13x + 6$ , find all the zeros of $P$ .

Separate answers with commas, e.g. 2, -5

Practice 8

The polynomial $P(x) = x^3 + ax^2 + bx - 6$ has $x - 1$ as a factor, and it leaves a remainder of $-8$ when divided by $x + 1$ . Find $(a, b)$ .

Enter a point like (2, -3)