Dividing Polynomials & the Factor Theorem — Print
Dividing polynomials is the long division you already know, and a remainder of zero names a factor.
What the practice questions cover
These questions use 7 ideas from the course. Follow any of them for a definition, where it is taught, and what it leads to.
- Rational — Any number writable as ab with integers a and b, b ≠ 0.
- Factor — A number that divides evenly into another.
- Roots — The x values where y = 0.
- 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.
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.
43
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.
Problem 1
Find the remainder when f(x)=x2+3x-5 is divided by x-1.
Worked answer
-1
f(1)=1+3-5=-1
Problem 2
Find the remainder when f(x)=x3-2x+4 is divided by x-2.
Worked answer
8
f(2)=8-4+4=8
Problem 3
What value goes in the synthetic division box when dividing by x+5?
Worked answer
-5
x+5=0 gives x=-5
Problem 4
What value goes in the box when dividing by x-7?
Worked answer
7
x-7=0 gives x=7
Problem 5
If f(3)=0, which of these is a factor of f(x)?
- x-3
- x+3
- 3x
- x
Worked answer
x-3
A root at 3 means x-3 divides exactly
Problem 6
Divide x2+7x+12 by x+3.
Worked answer
x+4
x2 ÷ x = x
Multiply back and subtract to get 4x+12
4x ÷ x = 4, remainder 0
Problem 7
Divide x2-x-6 by x-3.
Worked answer
x+2
x2-x-6=(x-3)(x+2)
Problem 8
Divide 2x2+5x-3 by x+3.
Worked answer
2x-1
2x2 ÷ x = 2x
2x(x+3)=2x2+6x, subtract to get -x-3
-x ÷ x = -1, remainder 0
Problem 9
Is x-2 a factor of x3-3x2+4? Answer yes or no.
Worked answer
yes
f(2)=8-12+4=0
A zero remainder means yes
Problem 10
Is x+1 a factor of x3+2x2-5? Answer yes or no.
Worked answer
no
f(-1)=-1+2-5=-4
Not zero, so no
Problem 11
Divide x3-8 by x-2.
Worked answer
x^2+2x+4
Synthetic division with 2: 1, 2, 4, remainder 0
Problem 12
Factor x3-7x+6 completely. Give the three roots, smallest first, separated by commas.
Worked answer
-3,1,2
f(1)=1-7+6=0
Dividing gives x2+x-6=(x+3)(x-2)
Roots -3, 1, 2
Problem 13
One root of x3-4x2+x+6 is -1. Find the other two, smallest first.
Worked answer
2,3
Dividing gives x2-5x+6
(x-2)(x-3)
Problem 14
For what value of k is x-2 a factor of x3+kx-6?
Worked answer
-1
8+2k-6=0
2k=-2 so k=-1
Problem 15
Find the remainder when 2x3-3x2+x-5 is divided by x+1.
Worked answer
-11
2(-1)-3(1)+(-1)-5
=-2-3-1-5=-11
Problem 16
Which list contains every possible rational root of x3-2x2+x-6?
- ±1,±2,±3,±6
- ±1,±2
- ±1,±6
- ±2,±3
Worked answer
±1,±2,±3,±6
The constant is -6
Its divisors are 1, 2, 3, 6, positive and negative