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§ 6.5
Synthetic Division and the Remainder
Theorem
Synthetic Division of Polynomials
We can use a process called synthetic division to divide
polynomials if the divisor is of the form x - c .
This method is just a shortcut for long division. With this process we can save both steps and paper by writing down only what is
necessary from the long division problem and by also compacting
the form.
Synthetic division is quick and can be extremely useful. Note that
this process works only when you can express the divisor in the
form x – c.
Blitzer, Algebra for College Students, 6e – Slide #2 Section 6.5
Synthetic Division of Polynomials
Synthetic Division
To divide a polynomial by x – c:
STEPS
1) Arrange polynomials in descending
powers, with a 0 coefficient for any missing
terms.
2) Write c for the divisor, x – c. To the
right, write the coefficients of the dividend.
3) Write the leading coefficient of the
dividend on the bottom row.
4) Multiply c (in this case, 3) times the
value just written on the bottom row. Write
the product in the next column in the
second row.
Blitzer, Algebra for College Students, 6e – Slide #3 Section 6.5
EXAMPLES
x  3 x3  4 x 2  5x  5
3 1 4 5 5
3 1 4 5 5
1
3 1 4 5 5
3
1
Synthetic Division of Polynomials
CONTINUED
Synthetic Division
To divide a polynomial by x – c:
STEPS
EXAMPLES
5) Add the values in this new column,
writing the sum in the bottom row.
3 1 4 5 5
3
1 7
6) Repeat this series of multiplications and
additions until all columns are filled in.
3 1 4 5 5
3 21 48
1 7 16 53
7) Use the numbers in the last row to write
the quotient, plus the remainder above the
divisor. The degree of the first term of
the quotient is one less than the degree of
the first term of the dividend. The final
value in this row is the remainder.
1x 2  7 x  16 
x  3 x3  4 x 2  5x  5
Blitzer, Algebra for College Students, 6e – Slide #4 Section 6.5
53
x3
Synthetic Division of Polynomials
EXAMPLE
Use synthetic division to divide x 2  6 x  6 x3  x 4 by 6  x.
SOLUTION
The divisor must be in the form x – c. Thus, we write 6 + x as
x – (-6). This means that c = -6. Rearranging the terms of the
dividend in descending order and writing a 0 for the missing
constant-term we can express the division as follows:
x   6 x 4  6 x3  x 2  6 x  0
Blitzer, Algebra for College Students, 6e – Slide #5 Section 6.5
Synthetic Division of Polynomials
CONTINUED
Now we are ready to set up the problem so that we can use
synthetic division.
6
This is c in
x – (-6).
1 6
1 6
0
Use the coefficients
of the dividend
x 4  6 x3  x 2  6 x  0
in descending
powers of x.
We begin the synthetic division process by bringing down 1.
This is followed by a series of multiplications and additions.
1) Bring down 1.
6
1 6
1
1 6
2) Multiply: -6(1) = -6.
0
6
1 6
6
1
Blitzer, Algebra for College Students, 6e – Slide #6 Section 6.5
1 6
0
Synthetic Division of Polynomials
CONTINUED
3) Add: -6 + -6 = -12.
6
1 6
6
1  12
1 6
4) Multiply: -6(-12) = 72.
0
1 6 1 6
 6 72
1  12 73
1 6 1 6
 6 72
1  12
0
6) Multiply: -6(73) = -438.
5) Add: 1 + 72 = 73.
6
6
0
6
1 6 1
6 0
 6 72  438
1  12 73
Blitzer, Algebra for College Students, 6e – Slide #7 Section 6.5
Synthetic Division of Polynomials
CONTINUED
7) Add: -6 + -438 = -444.
8) Multiply: -6(-444) = 2664.
6
6
1 6 1
6 0
 6 72  438
1  12 73  444
1 6 1
6 0
 6 72  438 2664
1  12 73  444
9) Add: 0 + 2664 = 2664.
6
1 6 1
6 0
 6 72  438 2664
1  12 73  444 2664
Blitzer, Algebra for College Students, 6e – Slide #8 Section 6.5
Synthetic Division of Polynomials
CONTINUED
The numbers in the last row represent the coefficients of the
quotient and the remainder. The degree of the first term of the
quotient is one less than that of the dividend. Because the degree
of the dividend, x 2  6 x  6 x 3  x 4, is 4, the degree of the quotient
is 3. This means that the 1 in the last row represents 1x 3 .
6
Thus,
1 6 1
6 0
 6 72  438 2664
1  12 73  444 2664
2664
x  12 x  73x  444 
x6
x  6 x 2  6 x  6 x3  x 4
3
2
Blitzer, Algebra for College Students, 6e – Slide #9 Section 6.5
The Remainder Theorem
The Remainder Theorem
If the polynomial f(x) is divided by x –c,
then the remainder is f(c).
This theorem is extremely useful because it can be used to evaluate a polynomial
function at c. Rather than having to substitute c for x in the function, you can just
divide the function by x – c using the shortcut of synthetic division. The remainder
will be the function value at c.
Blitzer, Algebra for College Students, 6e – Slide #10 Section 6.5
The Remainder Theorem
EXAMPLE
Given f x   4 x3  5x 2  6 x  4 , use synthetic division and the
Remainder Theorem to find f (-2).
SOLUTION
By the Remainder Theorem, if f (x) is divided by x – (-2) = x + 2,
then the remainder is f (2). We’ll use synthetic division to divide.
2
4
4
5
8
3
6
6
0
4
0
4
Remainder
The remainder, -4, is the value of f (-2). Thus, f (-2) = -4.
Blitzer, Algebra for College Students, 6e – Slide #11 Section 6.5
The Remainder Theorem
EXAMPLE
Show that -1 is a solution of the equation x 3  2 x 2  x  2  0.
Then solve the polynomial equation.
SOLUTION
One way to show that -1 is a solution is to substitute -1 for x in the
equation and obtain 0. Another way is to use synthetic division
and the Remainder Theorem.
1
1
1
Proposed
Solution
2
1
3
1
3
2
2
2
0
x 2  3x  2
x  1 x3  2 x 2  x  2
Remainder
Blitzer, Algebra for College Students, 6e – Slide #12 Section 6.5
The Remainder Theorem
CONTINUED
Equivalently,


x3  2 x 2  x  2  x  1 x 2  3x  2 .
Because the remainder is 0, the polynomial has a value of 0 when
x = -1. Thus, -1 is a solution of the given equation.
The synthetic division also shows that x + 1 is a factor of the
polynomial, as shown to the right of the synthetic division. The
other factor is the quotient found in the last row of the synthetic
division. Now we can solve the polynomial equation.
Blitzer, Algebra for College Students, 6e – Slide #13 Section 6.5
The Remainder Theorem
CONTINUED
x3  2 x 2  x  2  0
This is the given equation.
x  1x 2  3x  2  0
Factor using the result from
the synthetic division.
x 1x 1x  2  0
x 1  0
x  1
Factor the trinomial.
x 1  0
x2  0
x 1
x2
Set each factor equal to zero.
Solve for x.
The solutions are -1, 1, and 2, and the solution set is {-1,1,2}.
Blitzer, Algebra for College Students, 6e – Slide #14 Section 6.5