Algebra 2 · A.APR.D.6

Synthetic division

Long division by x minus a number writes every power of x over and over. How little of it do you actually need?

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Synthetic division

Factoring special patterns

When you reach for it

Long division of polynomials works for any divisor, but most of what it writes is repeated. When the divisor is linear, of the form x - c, nearly all of it can go: the powers of x, the copies of leading terms, and every subtraction. What is left is one row of coefficients, one number c, and a rhythm of multiply and add. That shorthand is synthetic division.

Reach for it whenever you divide a polynomial by x - c, x + c or ax - b, and above all to test many possible zeros quickly, which is how the rational root theorem is put to work.

The procedure

One division, done both ways

Divide x^3 + 2x^2 - 5x + 4 by x - 2 by long division, but write every line in columns, one column per power, keeping only the coefficients.

	x^3	x^2	x	1
dividend	1	2	-5	4
take away x^2(x - 2)	1	-2
what is left		4	-5
take away 4x(x - 2)		4	-8
what is left			3	4
take away 3(x - 2)			3	-6
remainder				10

The quotient is x^2 + 4x + 3 and the remainder is 10. Now look at what the table repeats.

The column headings already say which power each number belongs to, so the variables inside the division carry nothing. The first number of each “take away” row only cancels the leading number above it, which it always does, so it can go. The second number of each “take away” row is the quotient coefficient times -2, and it is taken away. Taking away -2 times a number is the same as adding 2 times it, so write 2, the number c in x - c, and add instead.

What survives is the dividend’s coefficients, the products 2, 8, 6, and the column sums:

\begin{array}{r|rrrr} 2 & 1 & 2 & -5 & 4 \\ & & 2 & 8 & 6 \\ \hline & 1 & 4 & 3 & 10 \end{array}

Figure 1Each arrow multiplies by 2, the number c in x - c; each column is then added down. The bottom row is the quotient’s coefficients, then the remainder.

The bottom row is the quotient’s coefficients and the remainder. Synthetic division is long division with the variables stripped out and each subtraction turned into an addition.

To fill one in, write c in the corner and the dividend’s coefficients along the top, highest power first. Bring the first coefficient straight down: the quotient’s leading coefficient equals the dividend’s, since x times it must make the leading term. Then, column by column, multiply the newest bottom number by c, write the product under the next coefficient, and add down. The last sum is the remainder. The numbers before it are the quotient’s coefficients, starting one power lower than the dividend, because dividing by x - c lowers the degree by one.

Why it works for every polynomial

Take a cubic a_3x^3 + a_2x^2 + a_1x + a_0. Dividing by x - c means finding a quadratic b_2x^2 + b_1x + b_0 and a number r with

\begin{aligned} &a_3x^3 + a_2x^2 + a_1x + a_0 \\ &= (x - c)(b_2x^2 + b_1x + b_0) + r \end{aligned}

Distributing on the right gives

\begin{aligned} &b_2x^3 + (b_1 - cb_2)x^2 \\ &+ (b_0 - cb_1)x + (r - cb_0) \end{aligned}

Two polynomials are equal for every x only when their matching coefficients are equal, so

\begin{aligned} b_2 &= a_3, & b_1 &= a_2 + cb_2, \\ b_0 &= a_1 + cb_1, & r &= a_0 + cb_0 \end{aligned}

Each equation says the same thing: the next bottom number is the coefficient above it plus c times the bottom number before it. That is the tableau, step for step. Nothing used the degree being 3 or c being a whole number, so the method works for any degree and any number c.

Dividing by x + c and by ax - b

The tableau needs the divisor written as x - c. For x + 2, write x - (-2): the corner number is -2.

For a divisor such as 2x - 1, whose x has coefficient 2, factor it as \displaystyle 2\left(x - \frac{1}{2}\right). Dividing p(x) by \displaystyle x - \frac{1}{2} gives \displaystyle p(x) = \left(x - \frac{1}{2}\right)Q(x) + r, and moving the factor 2 across,

p(x) = (2x - 1)\cdot\frac{Q(x)}{2} + r

So the quotient by 2x - 1 is Q divided by 2, and the remainder is r, unchanged. In general, divide by \displaystyle x - \frac{b}{a}, then divide the quotient, not the remainder, by a.

Missing powers

Every column stands for one power, so a missing power still needs its column, holding 0, just as long division needs placeholders.

Worked examples

Example 1

A placeholder zero

Divide 2x^4 - 5x^3 + x - 7 by x - 3.

Solution

The divisor is x - 3, so c = 3. There is no x^2 term, so the coefficients are 2, -5, 0, 1, -7, with 0 holding the x^2 column.

\begin{array}{r|rrrrr} 3 & 2 & -5 & 0 & 1 & -7 \\ & & 6 & 3 & 9 & 30 \\ \hline & 2 & 1 & 3 & 10 & 23 \end{array}

Bring down 2. Then 2 \cdot 3 = 6 and -5 + 6 = 1; 1 \cdot 3 = 3 and 0 + 3 = 3; 3 \cdot 3 = 9 and 1 + 9 = 10; 10 \cdot 3 = 30 and -7 + 30 = 23.

The dividend has degree 4, so the quotient has degree 3: it is 2x^3 + x^2 + 3x + 10, with remainder 23. As one statement,

\begin{aligned} \frac{2x^4 - 5x^3 + x - 7}{x - 3} &= 2x^3 + x^2 + 3x + 10 \\ &\quad + \frac{23}{x - 3} \end{aligned}

Check: (x - 3)(2x^3 + x^2 + 3x + 10) = 2x^4 - 5x^3 + x - 30, and adding 23 gives the dividend.

Example 2

Dividing by x + c

Divide x^3 + 4x^2 - 9 by x + 2.

Solution

Since x + 2 = x - (-2), the corner number is c = -2. The x term is missing, so the coefficients are 1, 4, 0, -9.

\begin{array}{r|rrrr} -2 & 1 & 4 & 0 & -9 \\ & & -2 & -4 & 8 \\ \hline & 1 & 2 & -4 & -1 \end{array}

Bring down 1. Then 1(-2) = -2 and 4 + (-2) = 2; 2(-2) = -4 and 0 + (-4) = -4; (-4)(-2) = 8 and -9 + 8 = -1.

The quotient is x^2 + 2x - 4 and the remainder is -1. Check: (x + 2)(x^2 + 2x - 4) = x^3 + 4x^2 - 8, and -8 + (-1) = -9.

Example 3

Several placeholders

Divide x^5 - 32 by x - 2.

Solution

Here c = 2, and four powers are missing, so the coefficients are 1, 0, 0, 0, 0, -32.

\begin{array}{r|rrrrrr} 2 & 1 & 0 & 0 & 0 & 0 & -32 \\ & & 2 & 4 & 8 & 16 & 32 \\ \hline & 1 & 2 & 4 & 8 & 16 & 0 \end{array}

Each column adds 0 to twice the previous bottom number, so the bottom row doubles, until the last column, where -32 + 32 = 0. The quotient is x^4 + 2x^3 + 4x^2 + 8x + 16 and the remainder is 0, so the division is exact:

\begin{aligned} &x^5 - 32 \\ &= (x - 2)(x^4 + 2x^3 + 4x^2 + 8x + 16) \end{aligned}

Example 4

A divisor of the form ax - b

Divide 4x^3 - 2x^2 + 6x - 1 by 2x - 1.

Solution

Write \displaystyle 2x - 1 = 2\left(x - \frac{1}{2}\right) and divide by \displaystyle x - \frac{1}{2} first, with \displaystyle c = \frac{1}{2}:

\begin{array}{r|rrrr} \tfrac{1}{2} & 4 & -2 & 6 & -1 \\ & & 2 & 0 & 3 \\ \hline & 4 & 0 & 6 & 2 \end{array}

Then \displaystyle 4 \cdot \frac{1}{2} = 2 and -2 + 2 = 0; \displaystyle 0 \cdot \frac{1}{2} = 0 and 6 + 0 = 6; \displaystyle 6 \cdot \frac{1}{2} = 3 and -1 + 3 = 2. The bottom 0 is a real coefficient, so the quotient by \displaystyle x - \frac{1}{2} is 4x^2 + 0x + 6 = 4x^2 + 6, with remainder 2. Moving a factor 2 across, \left(x - \tfrac{1}{2}\right)(4x^2 + 6) = (2x - 1)(2x^2 + 3). So

\begin{aligned} &4x^3 - 2x^2 + 6x - 1 \\ &= (2x - 1)(2x^2 + 3) + 2 \end{aligned}

The quotient is 2x^2 + 3 and the remainder is 2. Only the quotient is halved, because only it is multiplied by the divisor. Check: (2x - 1)(2x^2 + 3) = 4x^3 - 2x^2 + 6x - 3, and -3 + 2 = -1.

Example 5

The slope of a secant line

For f(x) = 2x^3 - 5x^2 + x + 4, find the slope of the line through (2, f(2)) and (x, f(x)) as a polynomial in x, for any x \neq 2. Then find the slope for x = 3.

Solution

The slope is rise over run, \displaystyle \frac{f(x) - f(2)}{x - 2}, the average rate of change of f from 2 to x. Here f(2) = 16 - 20 + 2 + 4 = 2, so the rise is 2x^3 - 5x^2 + x + 2. Divide it by x - 2, with c = 2:

\begin{array}{r|rrrr} 2 & 2 & -5 & 1 & 2 \\ & & 4 & -2 & -2 \\ \hline & 2 & -1 & -1 & 0 \end{array}

The remainder is 0, so for every x \neq 2 the slope is exactly 2x^2 - x - 1. At x = 3 the slope is 18 - 3 - 1 = 14.

Check: f(3) = 54 - 45 + 3 + 4 = 16, and \displaystyle \frac{16 - 2}{3 - 2} = 14.

Example 6

A quadratic divisor, one linear factor at a time

Divide x^4 + 2x^3 - 7x^2 - 8x + 12 by x^2 + 2x - 3.

Solution

Synthetic division cannot take a quadratic divisor, but this one factors: x^2 + 2x - 3 = (x - 1)(x + 3). Divide by x - 1, then divide that quotient by x + 3.

\begin{array}{r|rrrrr} 1 & 1 & 2 & -7 & -8 & 12 \\ & & 1 & 3 & -4 & -12 \\ \hline & 1 & 3 & -4 & -12 & 0 \end{array}

So the dividend is (x - 1)(x^3 + 3x^2 - 4x - 12). Now divide the cubic by x + 3, with c = -3:

\begin{array}{r|rrrr} -3 & 1 & 3 & -4 & -12 \\ & & -3 & 0 & 12 \\ \hline & 1 & 0 & -4 & 0 \end{array}

So x^3 + 3x^2 - 4x - 12 = (x + 3)(x^2 - 4), and together

\begin{aligned} &x^4 + 2x^3 - 7x^2 - 8x + 12 \\ &= (x - 1)(x + 3)(x^2 - 4) \\ &= (x^2 + 2x - 3)(x^2 - 4) \end{aligned}

The quotient is x^2 - 4 and the remainder is 0. With remainders r_1 and then r_2, two steps still work: p = (x - 1)\big((x + 3)Q_2 + r_2\big) + r_1, so the quotient is Q_2 and the remainder is r_2(x - 1) + r_1, not either number alone. And x^2 - 4 is a difference of squares, so the dividend is (x - 1)(x + 3)(x - 2)(x + 2).

Boundaries

The divisor must be linear. The tableau runs on one number, c. Taking away a multiple of a divisor of degree 2 or more, such as x^2 + 1, changes columns two or more places along, which a single multiply and add cannot record; use long division. Example 6 worked only because that quadratic split into linear factors.

The coefficient of x must be 1. For ax - b, divide by \displaystyle x - \frac{b}{a}, then divide the quotient by a and leave the remainder alone. Skip that last step and the quotient is a times too big.

Every power needs its column, written highest first, with 0 for each missing term.

The remainder is a single number. A divisor of degree 1 always leaves a constant, which may be negative, a fraction or zero. Zero means the division is exact.

Where this goes next

The remainder in the last column says more than it seems to, and the remainder and factor theorems explain what it tells you about the polynomial at c. With that, synthetic division becomes a quick test for zeros: the rational root theorem lists the only fractions that could be zeros, synthetic division tries each one, and every success leaves a quotient one degree lower to keep working on. Those zeros are where a graph meets the x-axis, the subject of whether it crosses or touches there and of graphing a polynomial.

Exercises

Use synthetic division to find the quotient and remainder.

Dividing by x - c

(x^3 - 4x^2 + x + 6) \div (x - 2)
(3x^3 + x^2 - 5x + 2) \div (x - 1)
(x^2 + 7x - 5) \div (x - 4)
(2x^4 - x^3 + 3x^2 - 4x + 5) \div (x - 2)

Dividing by x + c

(x^3 + 6x^2 + 5x - 8) \div (x + 3)
(2x^3 - 3x^2 - 4x + 7) \div (x + 1)
(-x^3 + 2x^2 + 4x - 3) \div (x + 2)

Missing powers

(x^4 - 3x^2 + 2x + 1) \div (x - 1)
(x^3 + 27) \div (x + 3)
(2x^5 - x^2 + 4) \div (x - 1)
(x^4 + x^3 - x - 1) \div (x + 1)
(t^3 - 2t^2 - 9) \div (t - 3)

Divisors of the form ax - b

(6x^3 + 7x^2 - 4x + 5) \div (3x + 2)
(8x^3 - 2x^2 + x - 3) \div (4x - 3)
(2x^3 + x^2 + 5) \div (2x + 3)

Using the result

Write 3x^3 - 2x^2 + 4 in the form (x - 2)q(x) + r.
Divide x^3 - 4x^2 - 7x + 10 by x - 1, divide that quotient by x + 2, and write x^3 - 4x^2 - 7x + 10 as a product of three linear factors.
For g(x) = x^3 + x^2 - 4x, find the slope of the line through (1, g(1)) and (x, g(x)) as a polynomial in x, for x \neq 1, and then its value at x = 4.

Reasoning

Explain why synthetic division by x - c writes c in the corner and adds, where long division takes away.
A polynomial of degree n is divided by x - c. How many numbers are in the bottom row, what is the degree of the quotient, and why is the remainder a constant?
Dividing 4x^3 - 6x^2 + 2x + 10 by 2x - 1 with \displaystyle c = \frac{1}{2} gives the bottom row 4, -4, 0 \mid 10. Show that (2x - 1)(2x^2 - 2x) + 10 equals the dividend, and explain why 4x^2 - 4x is not the quotient by 2x - 1.
Explain why synthetic division cannot divide by x^2 - 3 in one tableau.
Divide x^3 - 1 and x^5 - 1 by x - 1. Use the tableau to say what (x^n - 1) \div (x - 1) gives for any whole number n \ge 1, and why.

Answers

Quotient x^2 - 2x - 3, remainder 0. With c = 2 the bottom row is 1, -2, -3 \mid 0.
Quotient 3x^2 + 4x - 1, remainder 1. With c = 1: 3, 4, -1 \mid 1.
Quotient x + 11, remainder 39. With c = 4: 1, 11 \mid 39.
Quotient 2x^3 + 3x^2 + 9x + 14, remainder 33. With c = 2: 2, 3, 9, 14 \mid 33.
Quotient x^2 + 3x - 4, remainder 4. With c = -3: 1, 3, -4 \mid 4.
Quotient 2x^2 - 5x + 1, remainder 6. With c = -1: 2, -5, 1 \mid 6.
Quotient -x^2 + 4x - 4, remainder 5. With c = -2: -1, 4, -4 \mid 5.
Quotient x^3 + x^2 - 2x, remainder 1. Top row 1, 0, -3, 2, 1 with c = 1; bottom row 1, 1, -2, 0 \mid 1, where the 0 is the quotient’s constant term.
Quotient x^2 - 3x + 9, remainder 0. Top row 1, 0, 0, 27 with c = -3; bottom row 1, -3, 9 \mid 0.
Quotient 2x^4 + 2x^3 + 2x^2 + x + 1, remainder 5. Top row 2, 0, 0, -1, 0, 4 with c = 1; bottom row 2, 2, 2, 1, 1 \mid 5.
Quotient x^3 - 1, remainder 0. Top row 1, 1, 0, -1, -1 with c = -1; bottom row 1, 0, 0, -1 \mid 0.
Quotient t^2 + t + 3, remainder 0. Top row 1, -2, 0, -9 with c = 3; bottom row 1, 1, 3 \mid 0.
Quotient 2x^2 + x - 2, remainder 9. With \displaystyle c = -\frac{2}{3} the bottom row is 6, 3, -6 \mid 9; divide the quotient, not the remainder, by 3.
Quotient 2x^2 + x + 1, remainder 0. With \displaystyle c = \frac{3}{4}: 8, 4, 4 \mid 0; divide the quotient by 4.
Quotient \displaystyle x^2 - x + \frac{3}{2}, remainder \displaystyle \frac{1}{2}. With \displaystyle c = -\frac{3}{2}: \displaystyle 2, -2, 3 \mid \frac{1}{2}; divide the quotient by 2.
3x^3 - 2x^2 + 4 = (x - 2)(3x^2 + 4x + 8) + 20. Top row 3, -2, 0, 4 with c = 2; bottom row 3, 4, 8 \mid 20.
(x - 1)(x + 2)(x - 5). With c = 1: 1, -3, -10 \mid 0, so the first quotient is x^2 - 3x - 10. With c = -2 on that: 1, -5 \mid 0, which leaves x - 5.
Slope x^2 + 2x - 2, which is 22 at x = 4. Since g(1) = -2, divide x^3 + x^2 - 4x + 2 by x - 1: with c = 1 the bottom row is 1, 2, -2 \mid 0.
Each step of long division takes away a multiple q(x - c), whose entry in the next column is -cq. Taking away -cq is the same as adding cq. Writing c rather than -c in the corner turns every subtraction into an addition, so the whole method is multiply by c and add.
The top row has n + 1 coefficients, so the bottom row has n + 1 numbers: n quotient coefficients and the remainder. The quotient has degree n - 1, since x times its leading term must give x^n. The remainder is a constant because it must have lower degree than the divisor, and x - c has degree 1.
The quotient by 2x - 1 is 2x^2 - 2x and the remainder is 10. Expanding, (2x - 1)(2x^2 - 2x) = 4x^3 - 6x^2 + 2x, and adding 10 gives the dividend. The row 4, -4, 0 belongs to the divisor \displaystyle x - \frac{1}{2}, which is half of 2x - 1, so its quotient 4x^2 - 4x is twice too big.
Every step of synthetic division records taking away a multiple of the divisor with one number, c, which changes only the next column. Taking away a multiple of x^2 - 3 changes the column two places along, and a quadratic divisor in general changes two columns at once, so one multiply and add per column cannot record it. Long division can.
x^3 - 1 = (x - 1)(x^2 + x + 1) and x^5 - 1 = (x - 1)(x^4 + x^3 + x^2 + x + 1), remainder 0 each time. In general the quotient is x^{n-1} + x^{n-2} + \dots + x + 1 with remainder 0. With c = 1 and top row 1, 0, \dots, 0, -1, each column adds 1 times the previous bottom number, 1, to a 0, so every bottom number is 1 until the last column, where -1 + 1 = 0.