Division always has four parts
When you divide 17 by 5 you get 3 and 2 is left. We write 17 = 5 × 3 + 2. Here 17 is the dividend (the number being shared), 5 is the divisor (what we divide by), 3 is the quotient (the answer) and 2 is the remainder (what is left).
Polynomials follow the same rule. If we divide p(x) by a divisor d(x), we get a quotient q(x) and a remainder r(x):
p(x) = d(x) × q(x) + r(x), and the degree of r(x) is smaller than the degree of d(x).
When the divisor is a straight-line expression like (x − a) (degree 1), the remainder has degree 0. That means the remainder is just a number.
The remainder theorem
Remainder theorem: if p(x) is divided by (x − a), the remainder is p(a).
Why it is true
Write p(x) = (x − a) × q(x) + r. This is true for every x. Now choose x = a. The bracket (a − a) is 0, and 0 times anything is 0. So p(a) = 0 + r = r.
Picture it
On the graph of y = p(x), the remainder is the height of the curve above (or below) the point x = a. In the 3D scene the yellow bar shows this height.
Dividing by (ax − b)
If the divisor is (2x − 1), it becomes 0 when x = 1/2. So the remainder is p(1/2). In general, divide by (ax − b) and the remainder is p(b/a).
Watch the sign
- Divide by (x − 4): use x = 4.
- Divide by (x + 4): use x = −4, because x + 4 = x − (−4).
The factor theorem
A factor divides exactly, with nothing left over. So:
Factor theorem: (x − a) is a factor of p(x) if and only if p(a) = 0.
“If and only if” means it works both ways: if p(a) = 0 then (x − a) is a factor, and if (x − a) is a factor then p(a) = 0.
A value a with p(a) = 0 is called a zero (or root) of the polynomial. On the graph, it is a point where the curve meets the x-axis.
Where to search for zeros
If all the coefficients are whole numbers, any whole-number zero must divide the constant term exactly. For x³ − 2x² − 5x + 6, try the divisors of 6: ±1, ±2, ±3, ±6. This is called the rational root test (for fractions p/q, p divides the constant term and q divides the leading coefficient).
Long division and synthetic division
The remainder theorem gives only the remainder. To get the quotient too, divide.
Long division
Same as with numbers: divide the first term, multiply back, subtract, bring down, repeat.
Synthetic division (a quick way for x − a)
Write the coefficients in a row. Bring down the first one. Multiply by a, add to the next coefficient, and repeat. The last number is the remainder; the others are the quotient’s coefficients.
Example: (x³ − 2x² − 5x + 6) ÷ (x − 1), coefficients 1, −2, −5, 6 and a = 1:
- Bring down 1.
- 1 × 1 = 1; −2 + 1 = −1.
- −1 × 1 = −1; −5 + (−1) = −6.
- −6 × 1 = −6; 6 + (−6) = 0 ← remainder.
So the quotient is x² − x − 6 and the remainder is 0. Always write 0 for a missing power (x³ + 5 has coefficients 1, 0, 0, 5).
Try it: a practical
In the 3D: in the last step, move the slider for a. Before you let go, predict the remainder by working out p(a) on paper. Then check the yellow bar. Find all three values where the ball turns green.
At home: take 23 coins. Put them in piles of 4. Count full piles and coins left over. Then check: 4 × piles + left over = 23. You have just done dividend = divisor × quotient + remainder.
Key formulas and definitions
- Division: p(x) = d(x) · q(x) + r(x), degree of r < degree of d
- Remainder theorem: p(x) ÷ (x − a) leaves remainder p(a)
- p(x) ÷ (ax − b) leaves remainder p(b/a)
- Factor theorem: (x − a) is a factor ⇔ p(a) = 0
- Whole-number zeros divide the constant term
Worked examples
1. Find the remainder when x³ − 2x² − 5x + 6 is divided by (x − 2).
Put x = 2: 8 − 8 − 10 + 6 = −4. The remainder is −4.
2. Find the remainder when 2x³ + x² − 4 is divided by (x + 1).
x + 1 = 0 gives x = −1. p(−1) = −2 + 1 − 4 = −5. Remainder −5.
3. Find the remainder when 4x² − 2x + 3 is divided by (2x − 1).
2x − 1 = 0 gives x = 1/2. p(1/2) = 4·(1/4) − 1 + 3 = 1 − 1 + 3 = 3. Remainder 3.
4. Show that (x − 3) is a factor of x³ − 2x² − 5x + 6.
p(3) = 27 − 18 − 15 + 6 = 0. The remainder is 0, so by the factor theorem (x − 3) is a factor.
5. Find k if (x − 2) is a factor of x³ + kx² − 4x + 4.
p(2) = 0: 8 + 4k − 8 + 4 = 0, so 4k + 4 = 0 and k = −1.
6. Factorise x³ − 2x² − 5x + 6 fully.
Test divisors of 6: p(1) = 1 − 2 − 5 + 6 = 0, so (x − 1) is a factor. Synthetic division gives x² − x − 6 = (x − 3)(x + 2). So p(x) = (x − 1)(x − 3)(x + 2).
7. When p(x) is divided by (x − 1) the remainder is 3, and by (x + 2) the remainder is −6. Find the remainder when p(x) is divided by (x − 1)(x + 2).
The divisor has degree 2, so the remainder is ax + b. p(1) = a + b = 3 and p(−2) = −2a + b = −6. Subtract: 3a = 9, a = 3, b = 0. Remainder 3x.
Common mistakes
- Using the wrong sign: for (x + 5) you must put x = −5, not 5.
- For (2x − 3) putting x = 3 instead of x = 3/2. Set the divisor equal to 0 and solve.
- Forgetting a 0 for a missing power in synthetic division (x³ − 7x + 6 is 1, 0, −7, 6).
- Thinking a non-zero remainder means you made a mistake. It just means (x − a) is not a factor.