Synthetic division - Topics in precalculus (2024)

Synthetic division - Topics in precalculus (1)

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The remainder theorem

The factor theorem

IN ARITHMETIC we write, for example,

Synthetic division - Topics in precalculus (2)

or,

47
5
=9 + 2
5
.

Equivalently,

47 = 9·5 + 2

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

Dividend
Divisor
=Quotient + Remainder
Divisor

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

Synthetic division - Topics in precalculus (3)
x − 2
= x2 − 3x − 3 − 13
x − 2

Or,

Synthetic division - Topics in precalculus (4) =(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:

Synthetic division - Topics in precalculus (5)
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:

Synthetic division - Topics in precalculus (6)

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

Synthetic division - Topics in precalculus (7)

4. Add:

Synthetic division - Topics in precalculus (8)

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

Synthetic division - Topics in precalculus (9)

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 xa.

x + 3 = x − (−3).

Therefore, a = −3.

Here is the synthetic division:

Synthetic division - Topics in precalculus (10)

This tells us

Synthetic division - Topics in precalculus (11)
x + 3
= 2x4 − 3x3 + 9x2 − 2x + 6 19
x + 3

Or,

2x5 + 3x4 + 25x2 − 1 = (2x4 − 3x3 + 9x2 − 2x + 6)(x + 3) − 19.
Dividend = Quotient·Divisor + Remainder.

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.

To see the answer, pass your mouse over the colored area.
To cover the answer again, click "Refresh" ("Reload").

Synthetic division - Topics in precalculus (12)

x3 − 8x2 + x + 2 = (x2x − 6)(x − 7) − 40

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 xa.

That is, when

P(x) = Q(x)(xa) + R,

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

P(a) = R.

For,

P(a) = Q(a)(aa) + R
= Q(a)·0 + R
= 0 + R
= R.

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

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

Synthetic division - Topics in precalculus (13)

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:

Synthetic division - Topics in precalculus (14)

What does the remainder theorem tell us here?

f(5) = 0.

But this means that 5 is a root of f(x)Synthetic division - Topics in precalculus (15)

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.xr 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.

Synthetic division - Topics in precalculus (16)

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

Since the remainder is not 0 -- f(7) Synthetic division - Topics in precalculus (17)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.

Synthetic division - Topics in precalculus (18)

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.

Synthetic division - Topics in precalculus (19)

x3 + 125 = (x2 − 5x + 25)(x + 5)

Synthetic division - Topics in precalculus (20)

Next Topic: Roots of polynomials


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Synthetic division - Topics in precalculus (2024)

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