The remainder theorem

Dividing polynomials

When we divide $17$ by $5$, we get a quotient $3$ and a remainder $2$, and we write $17 = 5 \times 3 + 2$. In the same way, we can divide a polynomial by another polynomial.

The remainder has a smaller degree than the divisor. When the divisor is of degree $1$, such as $x - 3$, the remainder is a plain number.

Long division of polynomials takes time. Often we need only the remainder, and not the whole quotient. For that there is a shortcut.

The remainder theorem

If a polynomial $p(x)$ is divided by $x - a$, the remainder is $p(a)$.

Why? Divide by $x - a$. The remainder is a number, because its degree is less than $1$. Call it $r$. Then $p(x) = (x - a) \cdot q(x) + r$. Put $x = a$. The first term becomes zero, so $p(a) = r$.

Example 1. Find the remainder when $p(x) = x^3 + 3x^2 + 3x + 1$ is divided by $x + 2$.

We write $x + 2 = x - (-2)$, so $a = -2$.

The remainder is $-1$. We did not need to divide.

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

$a = 1$, so the remainder is $p(1) = 1 + 1 - 2 + 1 + 1 = 2$.

Example 3. A small check with numbers. Divide $x^2 + 3x + 4$ by $x - 1$ by long division. The quotient is $x + 4$ and the remainder is $8$. And $p(1) = 1 + 3 + 4 = 8$. The theorem agrees.

Dividing by ax + b

For a divisor $ax + b$, the number to use is $-\dfrac{b}{a}$, which is the zero of the divisor. The remainder is $p\left(-\dfrac{b}{a}\right)$.

Example: divide $p(x) = x^2 + 5x + 4$ by $2x + 1$. The zero of $2x + 1$ is $-\dfrac{1}{2}$. The remainder is $p\left(-\dfrac{1}{2}\right) = \dfrac{1}{4} - \dfrac{5}{2} + 4 = \dfrac{7}{4}$.

When the remainder is zero

If $p(a) = 0$, the remainder is $0$, and $x - a$ divides $p(x)$ exactly. So if $a$ is a zero of $p(x)$, then $x - a$ is a factor of $p(x)$. For example, $p(x) = x^2 - 5x + 6$ has $p(2) = 0$, so $x - 2$ is a factor. Indeed $x^2 - 5x + 6 = (x - 2)(x - 3)$.

This is a strong tool. To factorise a cubic, we try numbers until we find a zero. That zero gives one factor, and the rest is a quadratic that we can factorise as usual.

Move $a$ to $-2$, $1$ and $3$. At each one the remainder is $0$, so $x + 2$, $x - 1$ and $x - 3$ are the three factors.

Synthetic division

We can also divide by $x - a$ with a quick method. Write the coefficients, bring down the first one, and for each next coefficient add $a$ times the last result. The last number is the remainder. The others are the coefficients of the quotient.

For $x^3 + 3x^2 + 3x + 1$ divided by $x - 1$: bring down $1$. Then $3 + 1 \times 1 = 4$. Then $3 + 1 \times 4 = 7$. Then $1 + 1 \times 7 = 8$. The quotient is $x^2 + 4x + 7$ and the remainder is $8$, which equals $p(1) = 8$.

Practise with new numbers

Try it in Python

This program divides $x^3 + 3x^2 + 3x + 1$ by $x - a$ for four values of $a$, and it compares the remainder with $p(a)$.

Common mistakes

• Using the wrong sign. For $x + 2$, use $a = -2$, not $2$.
• Dividing by $2x + 1$ and using $a = 1$. Use $-\dfrac{1}{2}$.
• Forgetting that the remainder is a number when the divisor has degree $1$.
• Thinking that a remainder of $0$ means no answer. It means that the divisor is a factor.

Remember these

• Dividing $p(x)$ by $x - a$ leaves the remainder $p(a)$.
• For $x + 2$, use $a = -2$.
• We can find a remainder by putting a number in the polynomial, with no division.
• If $p(a) = 0$, then $x - a$ is a factor of $p(x)$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

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

2. What is $a$ for the divisor $x + 5$?

3. Is $x - 2$ a factor of $x^3 - 8$?

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