Synthetic division - Topics in precalculus (2024)

12

The remainder theorem

The factor theorem

IN ARITHMETIC we write, for example,

or,

Equivalently,

47 = 9·5 + 2

5 is called the divisor, 47 is the dividend, 9 is the quotient, and 2 is the remainder.

Or,

Dividend=Quotient·Divisor + Remainder.

In algebra, if we divide a polynomial P(x) by a polynomial D(x) (where the degree of D is less than the degree of P), we would find

P(x) = Q(x)·D(x) + R(x).

P(x) is the dividend, Q(x) is the quotient, and R(x) is the remainder.

For example, if, by long division, we divided

x³ − 5x² + 3 x − 7 by x − 2,

we would find

x − 2

=

x2 − 3x − 3 −

13 x − 2

Or,

=(x² − 3x − 3)(x − 2) − 13.

x³ − 5x² + 3x − 7 is the dividend, x² − 3x −3 is the quotient, and −13 is the remainder.

Here is how to do this problem by synthetic division.

First, to use synthetic division, the divisor must be of the first degree and must have the form x − a. In this example, the divisor is x − 2, with a=2.

Here again is the problem:

x − 2

Proceed as follows:

1. Write the coefficients of the dividend:

1 − 5 + 3 − 7

2. Put a, in this case 2, in a box to the right, leave a space, and draw a
2. line:

3. Bring down the leading coefficient (1), multiply it with a (2), and
3. write that product (1· 2) in the second column:

4. Add:

5. Repeat the process. −3·2 = −6. And so on, until all the coefficients
5. have been exhausted.

The first three numbers, 1 − 3 − 3, are the coefficients of the quotient, and the final number, −13, is the remainder.

We have

x³ − 5x² + 3 x − 7 = (x² − 3x −3)(x − 2) − 13.

Example 1.Use synthetic division to divide

2x⁵ + 3x⁴ + 25x² − 1 by x + 3.

Solution.There are a couple of points here. First, we must account for all six coefficients of the general form.

2 + 3 + 0 + 25 + 0 − 1

The coefficient of x³ is 0, as is the coefficient of x.

Next, the divisor is x + 3. But the divisor must have the form x − a.

x + 3 = x − (−3).

Therefore, a = −3.

Here is the synthetic division:

This tells us

x + 3

=

2x4 − 3x3 + 9x2 − 2x + 6 −

19 x + 3

Or,

Note: The degree of the quotient is one less than the degree of the dividend. And the degree of the remainder is less than the degree of the divisor, x + 3, which in this case is 1. The remainder therefore is of degree 0, which is a number.

In general, if we divide a polynomial of degree n by a polynomial of degree 1, then the degree of the quotient will be n − 1. And the remainder will be a number.

Problem 1.Use synthetic division to divide

x³ − 8x² + x + 2 by x − 7.

Write your answer in the form

P(x) = Q(x)·D(x) + R.

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The remainder theorem

The value of a polynomial P(x) at x = a,

P(a),

is equal to the remainder upon dividing P(x)
by x − a.

That is, when

P(x) = Q(x)(x − a) + R,

where Q(x) is the quotient and R is the remainder, then

P(a) = R.

For,

Example 2.Let f(x) = x³ − 3x² − 13x + 15.

We will use synthetic division to divide f(x) by x + 4.

Now, what does the remainder theorem tell us?

The value of f(x) at x = −4, is equal to the remainder:

f(−4) = −45.

Now let us divide f(x) by x − 5:

What does the remainder theorem tell us here?

f(5) = 0.

But this means that 5 is a root of f(x)

Moreover, since the remainder is 0 -- there is no remainder -- then (x−5) is a factor of f(x). The synthetic division shows:

x³ − 3x² − 13x + 15 = (x² + 2x − 3)(x − 5)

This illustrates the Factor Theorem:

The Factor Theorem.x − r is a factor of a polynomial P(x) if and only if r is a root of P(x).

Problem 2.Let f(x) = x³ − 5x² − 4x + 7. Use synthetic division to divide f(x) by x − 7.

Therefore, according to the remainder theorem, f(7) = 77.

Since the remainder is not 0 -- f(7) 0 -- upon dividing f(x) by x − 7, then (x − 7) is not a factor of f(x). And according to the factor theorem, 7 is not a root of f(x).

Problem 3.Let g(x) = 3x ⁴ + 17x³ + 16x ² − 10x + 4. Use synthetic division to divide g(x) by x + 2.

According to the remainder theorem, g(−2) = 0.

Therefore, what do you conclude about −2?

−2 is a root of g(x).

What do you conclude about (x + 2)?

(x + 2) is a factor of g(x).

Problem 4.Use synthetic division to divide

x³ + 125 by x + 5.

Next Topic: Roots of polynomials

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