Dividing Polynomials & the Factor Theorem

Dividing polynomials is the long division you already know, and a remainder of zero names a factor.

Does 37 divide 1554 exactly?

You would find out by long division. Bring down, multiply back, subtract, repeat.

It comes out 42 with nothing left over, and that is what makes 37 a factor of 1554.

Now ask whether x - 3 divides x² + x - 12. The layout is identical, and again nothing is left over.

A remainder of zero means "divides exactly" in both worlds. So division is a factor test — and it works on polynomials nobody could factor by inspection.

Long division, term by term

Write both polynomials in descending order, and put in a 0 for any missing power — a gap will throw the columns out.

Divide the leading terms, write the result on top, multiply back, subtract, bring the next term down.

Stop when what remains has a lower degree than the divisor. That leftover is the remainder.

Subtract the whole line

You are subtracting a bracket, so every sign in it flips. Forgetting the second sign is the most common error in this unit, and it derails everything after it.

Synthetic division, when the divisor is x - a

If you are dividing by x - a, the xs carry no information, so drop them and work with coefficients alone.

Write a in the box, bring down the first coefficient, multiply by a, add to the next, and repeat.

The last number is the remainder; the rest are the coefficients of the quotient, one degree lower.

Mind the sign of a

Dividing by x - 3 puts +3 in the box; dividing by x + 3 puts -3. Use the root, not the number you see.

The remainder theorem

There is a shortcut worth having: the remainder on dividing f(x) by x - a is simply f(a).

So to find the remainder for f(x) = x³ - 2x + 1 divided by x - 2, just compute f(2) = 5. No division needed.

It works because the division statement f(x) = (x-a)q(x) + r becomes f(a) = 0 · q(a) + r when you substitute x = a.

One substitution beats one division

When a question asks only for the remainder, evaluating f(a) is faster and far less error-prone than carrying out the division.

The factor theorem

If the remainder is zero then the divisor went in exactly, so x - a is a factor of f(x).

That gives a test with no guessing: f(a) = 0 if and only if x - a is a factor.

It also runs the other way. Knowing a root hands you a factor, and dividing by it leaves a smaller polynomial to finish.

Roots and factors are the same fact

"3 is a root", "f(3) = 0", "x - 3 is a factor" and "the graph crosses at 3" are four ways of saying one thing.

Factoring a cubic completely

For a cubic, test the small whole numbers that divide the constant term — those are the only rational roots worth trying.

When one works, divide it out. You are left with a quadratic, which you already know how to factor.

So a cubic takes one lucky guess and one routine step, not a flash of insight.

Where to guess

For x³ - 4x² + x + 6, try ±1, ±2, ±3, ±6 — the divisors of 6. Anything else cannot be a rational root.

Worked examples

Long division

Divide x³ + 3x² - 4x - 12 by x - 2.

x³ ÷ x = x². Multiply back: x³ - 2x², and subtract to get 5x². Subtracting the bracket flips both signs.
Bring down -4x. Now 5x² ÷ x = 5x; multiply back and subtract to get 6x. Same three moves, one degree lower.
Bring down -12. Then 6x ÷ x = 6; multiply back and subtract to get 0. Remainder 0, so x-2 is a factor.

x² + 5x + 6, remainder 0

A remainder without dividing

Find the remainder when f(x) = 2x³ - 5x + 4 is divided by x - 3.

The divisor is x - 3, so a = 3. Use the root, not the -3 on the page.
Compute f(3) = 2(27) - 5(3) + 4 = 54 - 15 + 4. The remainder theorem says this is the remainder.
That is 43. Not zero, so x - 3 is not a factor.

Factoring a cubic

Factor x³ - 4x² + x + 6 completely.

Test divisors of 6. f(-1) = -1 - 4 - 1 + 6 = 0, so x + 1 is a factor. The factor theorem turns a root into a factor.
Divide by x + 1 to get x² - 5x + 6. Synthetic division with -1 in the box.
Factor the quadratic: (x-2)(x-3). Two numbers multiplying to 6 and adding to -5.

(x+1)(x-2)(x-3)

Words to know

Dividend: The polynomial being divided — the one inside the bracket.
Divisor: The polynomial you are dividing by.
Remainder: What is left when the divisor no longer fits. Always lower degree than the divisor.
Remainder theorem: Dividing f(x) by x - a leaves remainder f(a).
Factor theorem: If f(a) = 0 then x - a is a factor, and the other way round.

Mistakes to watch for

Forgetting to subtract the whole line. Dividing x³ + 3x² by x-2 gives x³ - 2x² to subtract, and 3x² - (-2x²) is 5x², not x². Every sign in the bracket flips, and one missed sign wrecks the rest.

Leaving a gap for a missing power. In x³ - 8 the x² and x terms are missing, so write x³ + 0x² + 0x - 8. Without the placeholders the columns slide and the quotient comes out a degree wrong.

Putting the wrong number in the synthetic division box. Dividing by x + 4 uses -4, because -4 is the root. Using +4 answers a different question entirely.

Reading the last synthetic-division number as part of the quotient. It is the remainder. If a cubic divided by x-a leaves four numbers, the first three are the quotient and the fourth stands alone.

Guessing roots at random. Only divisors of the constant term can be rational roots, so for x³ - 4x² + x + 6 the whole list is ±1, ±2, ±3, ±6. Trying 5 is wasted effort.

Synthetic division, the shortcut with just the coefficients, was in use by the 1800s. It strips away the xs because they were never doing any work.

The remainder theorem says the remainder on dividing by x - a equals f(a). Checking a factor costs one substitution, not a whole division.

Computers factor enormous polynomials this way every day — it is how a graphing calculator finds roots it cannot see.

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