Dividing polynomials: long division, synthetic division, the remainder and factor theorems, and one cubic factored all the way

Dividing a polynomial by x − a does two jobs at once: the remainder tells you P(a) (the remainder theorem), and a remainder of zero tells you x − a is a factor (the factor theorem). Synthetic division is long division with the x's left out, and the one rule students break is the number in the box: dividing by x − 3 puts 3 in the box, not −3. Here is one cubic divided both ways, then factored completely, with the two faults named.

Why divide a polynomial at all

Dividing polynomials is how you factor anything past a quadratic. To factor a cubic, you find one root, divide out the matching linear factor, and factor the quadratic that remains. There is no shortcut formula a student is expected to use for a cubic, so the division is the route.

That is the whole plan for how to factor polynomials of degree three or higher:

1. Find one value a that makes the polynomial zero.
2. Divide by x − a.
3. Factor what is left.

In MHF4U these are idea 3, "Dividing polynomials and the remainder theorem", and idea 4, "The factor theorem and factoring polynomial expressions". Before them comes idea 2, "Graphs of polynomial functions: zeros, multiplicity and turning points". After them comes idea 5, "Solving polynomial equations". The MHF4U course map has the full order, and the precalculus ideas page explains why calculus keeps using this.

Polynomial long division, once

Here is the cubic for the whole page:

P(x) = 2x³ − 3x² − 11x + 6, divided by x − 3.

Polynomial long division works like the long division of numbers. You divide the leading terms, multiply back, subtract, and bring down.

1. Divide the leading terms. 2x³ ÷ x = 2x². Write 2x² in the quotient.
2. Multiply back. 2x²(x − 3) = 2x³ − 6x².
3. Subtract. (2x³ − 3x²) − (2x³ − 6x²) = 3x². Bring down the −11x to get 3x² − 11x.
4. Divide again. 3x² ÷ x = 3x. Write + 3x in the quotient.
5. Multiply back. 3x(x − 3) = 3x² − 9x.
6. Subtract. (3x² − 11x) − (3x² − 9x) = −2x. Bring down the + 6 to get −2x + 6.
7. Divide again. −2x ÷ x = −2. Write − 2 in the quotient.
8. Multiply back. −2(x − 3) = −2x + 6.
9. Subtract. (−2x + 6) − (−2x + 6) = 0.

The quotient is 2x² + 3x − 2 and the remainder is 0.

Check by multiplying back: (x − 3)(2x² + 3x − 2) = 2x³ + 3x² − 2x − 6x² − 9x + 6 = 2x³ − 3x² − 11x + 6. That is P(x).

Synthetic division, the same division line by line

Synthetic division keeps only the numbers. Every x is in the same column every time, so you do not need to write it. Here is how to do synthetic division on the same cubic.

1. The box. The divisor is x − 3, so the number in the box is 3. It is the root of the divisor, not the sign you see.
2. The coefficients. Write them in order of power, with a zero for any missing power: 2, −3, −11, 6.
3. Bring down the first coefficient: 2.
4. Multiply, then add. 2 × 3 = 6. Then −3 + 6 = 3.
5. Multiply, then add. 3 × 3 = 9. Then −11 + 9 = −2.
6. Multiply, then add. −2 × 3 = −6. Then 6 + (−6) = 0.
7. Read the bottom row. It is 2, 3, −2 | 0. The last number is the remainder, 0. The others are the quotient's coefficients, one degree lower than P: 2x² + 3x − 2.

That matches the long division exactly. Synthetic division is the same arithmetic with less writing.

The table, as rows of numbers:

• Box: 3
• Top row: 2, −3, −11, 6
• Middle row: (blank), 6, 9, −6
• Bottom row: 2, 3, −2 | 0

The common wrong line: the sign in the box

Students see x − 3 and write −3 in the box. Here is what happens when you do:

• Bring down 2.
• 2 × (−3) = −6; −3 + (−6) = −9.
• −9 × (−3) = 27; −11 + 27 = 16.
• 16 × (−3) = −48; 6 + (−48) = −42.

The bottom row is 2, −9, 16 | −42. The "remainder" is −42, and the work looks clean, which is what makes the fault hard to spot. But −42 is P(−3): you divided by x + 3, which is a different question. The box holds the value of x that makes the divisor zero. For x − 3, that value is 3.

Booth and colleagues (2013) found that studying incorrect examples, with the fault explained, may be especially useful for conceptual understanding in algebra. That is why the wrong row is written out in full.

The second wrong line: a missing power

Take x³ − 7x + 6. There is no x² term. A student who writes the coefficients as 1, −7, 6 has dropped a column. Divided by x − 1, that row gives 1, −6 | 0, which reads as the quotient x − 6 with remainder 0.

The zero remainder looks like success. It is not. A cubic divided by a linear factor must leave a quadratic, and x − 6 is linear. Those coefficients described x² − 7x + 6, a different polynomial. The correct row is 1, 0, −7, 6, with the zero holding the x² place.

The remainder theorem and the factor theorem

The two theorems are why division is worth doing.

The remainder theorem: when P(x) is divided by x − a, the remainder is P(a).

Check it on the example. P(3) = 2(27) − 3(9) − 11(3) + 6 = 54 − 27 − 33 + 6 = 0. That is the remainder both divisions found.

One more instance. The remainder on dividing P(x) by x + 1 is P(−1), because x + 1 = x − (−1):

P(−1) = 2(−1) − 3(1) − 11(−1) + 6 = −2 − 3 + 11 + 6 = 12.

Long division by x + 1 gives the quotient 2x² − 5x − 6 and the same remainder, 12.

The factor theorem: x − a is a factor of P(x) exactly when P(a) = 0.

So P(3) = 0 tells you (x − 3) is a factor before you divide anything. The division then hands you the other factor.

The common wrong line: the sign again

For the divisor x + 1, students evaluate P(1) and get −6. That is the remainder for x − 1, not x + 1. The rule is the same as the box: find the x that makes the divisor zero. x + 1 = 0 at x = −1, so you evaluate P(−1).

Finishing the factoring

The division left 2x² + 3x − 2. Factor it: 2x² + 3x − 2 = (2x − 1)(x + 2). Check: (2x − 1)(x + 2) = 2x² + 4x − x − 2 = 2x² + 3x − 2.

So the cubic, factored completely:

P(x) = (x − 3)(2x − 1)(x + 2), with zeros x = 3, x = 1/2 and x = −2.

How do you find the first root? The rational candidates are plus or minus a factor of the constant over a factor of the leading coefficient. Here that is ±1, ±2, ±3, ±6, ±1/2, ±3/2.

Test them with the remainder theorem, which is faster than dividing each time:

• P(1) = 2 − 3 − 11 + 6 = −6. Not a root.
• P(−1) = 12. Not a root.
• P(2) = 16 − 12 − 22 + 6 = −12. Not a root.
• P(−2) = −16 − 12 + 22 + 6 = 0. A root.

Tested in that order, −2 is the first hit, and it works just as well as 3. Synthetic division by x + 2 (box −2) gives 2, −7, 3 | 0, the quotient 2x² − 7x + 3 = (2x − 1)(x − 3), and the same three factors. Any root you find leads to the same answer. This page divided by x − 3 because that is where the example started.

Sibling to try cold: how to factor a cubic

Factor Q(x) = x³ − 7x + 6 completely. Then find the remainder when Q(x) is divided by x − 2.

Cover everything above and write every line yourself: the candidates, the test, the box, the coefficients with the zero, the quotient, the final factors. A worked example followed by your own attempt is the order Sweller and Cooper (1985) found teaches beginners in algebra faster than solving problems from the start.

Check

Q(x) = (x − 1)(x − 2)(x + 3). The remainder on dividing by x − 2 is 0.

If your quotient came out linear, you dropped the zero for the missing x² term.

When synthetic division does not apply

Synthetic division works only for a divisor of the form x − a: linear, with a leading coefficient of 1. For a divisor like x² + 1, use long division; the same divide, multiply back, subtract steps work with any divisor. For a divisor like 2x − 1, you can divide synthetically by x − 1/2 and then adjust. On our cubic, box 1/2 gives 2, −2, −12 | 0, so P(x) = (x − 1/2)(2x² − 2x − 12). Move the 2 across: P(x) = (2x − 1)(x² − x − 6). The quotient for 2x − 1 is x² − x − 6, half of the synthetic row.

Where it goes next

Solving polynomial equations and inequalities (ideas 5 and 6) is this factoring followed by a sign chart: once P(x) is in factors, you can read where it is zero, positive and negative. The oblique asymptote of a rational function is a polynomial division's quotient, worked on the asymptotes page.

Practise on paper with this page closed. Testing yourself before you reread explains why the closed page matters, and worked examples, then practice problems sets out the order. When the sign in the box keeps tripping you, chapter 4 of the free playbook is about finding the one line you keep getting wrong. The other MHF4U ideas are on the math help hub.

Dividing polynomials on one screen

• The box rule: the number in the box makes the divisor zero. x − 3 means 3; x + 1 means −1.
• Zeros for missing powers: x³ − 7x + 6 is written 1, 0, −7, 6.
• Synthetic division in four words: bring down, multiply, add, repeat.
• Remainder theorem: dividing P(x) by x − a leaves P(a).
• Factor theorem: P(a) = 0 exactly when x − a is a factor.
• The factoring route: test rational candidates, divide by the factor you found, factor the quadratic.
• Worked answer: 2x³ − 3x² − 11x + 6 = (x − 3)(2x − 1)(x + 2).