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Synthetic Division by x a

The remainder theorem

The factor theorem

IN ARITHMETIC we would 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.

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

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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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.

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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