Chapter 3 · Complex Numbers and Equations · Concept 2
3-15 Synthetic Division
Lesson 3-12 got the remainder in three lines but threw the quotient away. Long division keeps the quotient but takes fifteen lines. This keeps both, in four.
Guiding question
Is there a way to divide that is quick AND keeps the quotient?
Learning outcome. Use synthetic division to divide polynomials.
For dividing by x − a, and only that
When the divisor is a linear expression x − a, synthetic division finds the quotient and the remainder together, with nothing but multiplying and adding.
The five moves
- 1. Write the coefficients along the top — including a 0 for any missing power.
- 2. On the left, write the number that makes the divisor zero. For x + 2 that is −2.
- 3. Bring the first coefficient straight down.
- 4. Multiply it by the number on the left; write that under the next coefficient.
- 5. Add the column. Repeat 4 and 5 to the end.
Reading the bottom row
From the RIGHT: the last number is the remainder. Then the constant term of the quotient, then the coefficient of x, then of x², and so on.
Which is the same as reading it from the left as an ordinary polynomial, one degree lower than the one you started with — with the last number chopped off.
Two actions, taking turns. Press the button and follow the orange arrow — it always goes diagonally down and across, and it is always a multiplication.
dividing
x³ + 5x + 11 ÷ x + 2
Write the coefficients along the top, and the number that makes the divisor zero on the left.
Using synthetic division, find the quotient and remainder when x³ + 5x + 11 is divided by x + 2.
- 1
Take out the coefficients — and mind the gap.
1, 0, 5, 11There is no x² term, so its coefficient is 0. The book names leaving this out as the main mistake of the lesson.
Check it the way you would check any division.
(x + 2)(x² − 2x + 9) − 7 = x³ + 5x + 11 ✓
A = B × Q + R, exactly as in Chapter 1. If it does not rebuild, something in the table is wrong.
Work with a partner. One does it by synthetic division, the other by ordinary long division. You should land on the same quotient and the same remainder — and one of you will finish a lot sooner.
Two of these have gaps.
(a) has no x² term, and (d) has no x² term either. Their top rows are 1, 0, −5, 6 and 3, 5, 0, −2, −12. Count the numbers before you start: a polynomial of degree n needs n + 1 of them.
P(x) = 4x⁴ − 4x³ − 11x² + 12x − 3 crosses the x-axis at x = 1/2, with multiplicity 2. Find the other crossings.
Leaving out a zero for a missing term.
x³ + 5x + 11 goes in as 1, 0, 5, 11. The book names this as the main misconception, and it wrecks every column after it.
Mixing the remainder into the quotient.
The last number of the bottom row is the remainder and nothing else. Bottom row 1, −2, 9, −7 means x² − 2x + 9 with remainder −7.
Writing the wrong number on the left.
For x + 2 it is −2. Same rule as lesson 3-12: the number that makes the divisor zero.
Using it on a quadratic divisor.
This method is for x − a only. For x² + x − 6 you are back to long division, or to the remainder method of 3-13.
Not checking with A = B × Q + R.
One multiplication rebuilds the original and catches any slip in the table.
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