DIVIDING POLYNOMIALS; SYNTHETIC DIVISION
When doing long division, the steps are divide, multiply, subtract, bring down
Example 1
Divide x 2  7 x  6 by x  2 .
Solution. Arrange the terms in descending powers of x, then set up the division problem.
x  2 x2  7 x  6
Divide x 2 (the first term of the dividend) by x (the first term of the divisor):
x2
 x . Align like terms.
x
x
x  2 x2  7 x  6
Multiply each term in the divisor x  2 by x: xx  2  x 2  2 x . Align like terms of the product under like
terms in the dividend.
x
x  2 x  7x  6
2
x2  2 x
Subtract x 2  2 x from x 2  7 x by changing the sign of each term in the lower expression and adding.
x
x  2 x  7x  6
2
x2  2 x
5x
Bring Down the 6 from the original dividend and add algebraically to form a new dividend.
x
x  2 x2  7 x  6
x2  2 x
5x  6
Divide 5x (the first term of the new dividend) by x (the first term of the divisor):
5x
 5 . Align like terms.
x
x 5
x  2 x2  7 x  6
x2  2 x
5x  6
1
Multiply each term in the divisor x  2 by 5: 5x  2  5x  10 . Align like terms of the product under like
terms in the new dividend.
x 5
x  2 x  7x  6
2
x2  2 x
5x  6
5 x  10
Subtract 5x  10 from 5x  6 by changing the sign of each term in the lower expression and adding. The
remainder is  4 .
x 5
x  2 x  7x  6
2
x2  2 x
5x  6
5 x  10
4
4
x2  7 x  6
4
, which is also written as x  5 
.
 x5
x2
x2
x2
An important property of division can be illustrated by clearing fraction in the equation that concluded this
example. Multiplying both sides of this equation by x  2 results in the following:
 x2  7 x  6 
 4 

x  2  x  5x  2  
x  2
 x2
 x2 

x 2  7 x  6  x  5x  2  4
You can check your work by multiplying the divisor with the quotient and then adding the remainder. This
should give the dividend. This process is known as the Division Algorithm.
2
THE DIVISION ALGORITHM
If f x  and d x  are polynomials with d x  0 and the degree of d x  is less than the degree of f x  , then
there exists unique polynomials qx  and r x  such that
f x  =

Dividend
d x 
qx 


Divisor
r x 
+


Quotient
Remainder
The remainder, r x  , is either equal to zero, or is of degree less than th4e degree of d x  . If r x   0 , we
say that d x  divides evenly into f x  and that d x  and qx  are factors of f x  .
Example 2
Divide 4  5 x  x 2  6 x 3 by 3x  2 .
Missing Powers of x
In long division, if a power of x is missing in either a dividend or a divisor, add that power with a coefficient of
0 and then divide. In this way, like terms will be aligned as you carry out the long division. See below:
Example 3
Divide 6 x 4  5 x 3  3 x  5 by 3 x 2  2 x .
Synthetic Division
We can use a process called synthetic division to divide polynomials if the divisor is of the form x – c. This
method can be done more rapidly than using long division. Compare the two methods using the problem
below:
Long Division:
x  4 3x3  2 x 2
Synthetic Division:
 150
4
3
2
 0  150
Example 4
Use synthetic division to divide 5 x 3  6 x  8 by x  2 .
Your turn:
Use synthetic division to divide x 3  7 x  6 by x  2 .
3
The Remainder Theorem
Dividing a function f(x) by x – c produces a new function, q(x), and a remainder r.
That is,
f ( x)
r
 q( x) 
.
xc
xc
Multiplying both sides by x – c produces the following:
f ( x)  q( x)  x  c  r
Now let’s evaluate the function f(x) at the point where x = c:
f (c)  q(c)  c  c  r  q(c)  0  r  r
The result is known as the remainder theorem
REMAINDER THEOREM
If a polynomial f(x) is divided by x – c, then the remainder is f(c).
The Remainder Theorem can be used to evaluate polynomials.
Example 5
Given f ( x)  x 3  4 x 2  5x  3 , use the remainder theorem to evaluate f(2).
Your turn:
Given f ( x)  3x 3  4 x 2  5 x  3 , use the remainder theorem to evaluate f(-4).
Example 6
Use the Remainder Theorem to find the vertex of the parabola whose equation is
y  3x 2  6 x  4 .
The Factor Theorem
Look again at the division algorithm for the remainder theorem:
f ( x)
r
 q( x) 
xc
xc

f ( x)  q( x)  x  c  r
If the remainder is 0, then the procedure looks like this:
f ( x)
 q( x)
xc

f ( x)  q( x)  x  c
Look at the final result. This is equivalent to saying that if f(x) is divided by x – c with a remainder 0, then x – c
is a factor of f(x). Also, if we replace x with c , then we get f (c)  g (c)  c  c  g (c)  0  0 . This is called
the factor theorem
4
FACTOR THEOREM
If f x  is divided by x  c with a remainder 0, then x  c is a factor of f(x).
A polynomial f x  has a factor of x  c if and only if f c  0
The factor theorem tells us two things:
1. If f c  0 , then x  c is a factor of f x  .
2. If x  c is a factor of f x  , then f c  0 .
a)
Determine whether x  1 is a factor of 2 x 4  3x 2  5 x  7
b)
Determine whether x  1 is a factor of 3x5  2 x 4  x3  8 x 2  5 x  1
Example 8
a)
b)
Factor f ( x)  2 x3  3x 2  11x  6 given that 3 is a zero of f ( x) .
Solve the equation 2 x 3  3x 2  11x  6  0 .
Example 9
Solve the equation 15 x 3  14 x 2  3x  2  0 given that –1 is a zero of
f ( x)  15 x 3  14 x 2  3x  2 .
Example 7
Example 10 Show that x  2 and x  3 are factors of f x   2 x 4  7 x 3  4 x 2  27 x  18 . Then find the
remaining factors of f x  .
5
ZEROS OF POLYNOMIAL FUNCTIONS
Remember what it means to find the zeros of a polynomial function:
1. The zeros of a polynomial are the roots (or solutions) of the equation f ( x)  0 .
2. If the zeros do not have an imaginary solution, then those zeros indicate the x-intercepts of the graph of
the polynomial.
Example 1
Decide whether the number k is a zero of f ( x )
a)
f ( x)  x 3  4 x 2  9 x  6
k 1
b)
f ( x)  x 4  x 2  3x  1
k  4
There is a theorem that lists all possible rational zeros of a polynomial function. It is called the rational zero
theorem.
RATIONAL ZEROS THEOREM
If
p
p
is a rational number (in lowest terms), and if
is a zero of f ( x ) , a polynomial function with integer
q
q
coefficients, then p is a factor of the constant term, and q is a factor of the leading coefficient.
In order to use this theorem, here’s what we have to do:



List all the factors of the constant term.
List all the factors of the leading coefficient.
factors of the constant t erm
Possible rational zeros =
factors of the leading coefficien t
Example 2
List all possible rational zeros of f ( x)   x  3x  4 .
Solution:
The constant term is 4
4
2
List all factors of 4:
The leading coefficient is –1
List all factors of –1:
Possible rational zeros =
 1,  2,  4
factors of 4

 1,  2,  4 .
factors of  1
1
There are six possible rational zeros.
6
Your turn:
List all possible rational zeros of f ( x)  x  2 x  5 x  6 .
Example 3
List all possible rational zeros of f ( x)  15 x  14 x  3x  2 .
Solution:
The constant term is –2
3
2
3
2
List all factors of –2:
The leading coefficient is 15
List all factors of 15:
Possible rational zeros =
 1,  2
factors of - 2
1
2
1
2
1
2

 1,  2,  ,  ,  ,  ,  , 
factors of 15  1,  3,  5,  15
3
3
5
5 15 15
There are sixteen possible rational zeros. The actual zeros, as we can see from looking at a graph, are
 1,  1 3, 2 5,
which is 3 of the possible 16 zeros.
Your turn: List all possible rational zeros of f ( x)  4 x  12 x  x  3 .
Suppose that, instead of listing the rational zeros, we actually had to find the rational zeros of a polynomial
function. Here’s what we need to do:
5



4
Use trial-and-error to find one of the zeros using synthetic division.
Use the result of the synthetic division to factor the original polynomial.
Set each factor equal to zero to identify any additional rational zeros.
7
Example 4
Find all zeros of f ( x)  x  2 x  5 x  6 .
3
2
Possible rational zeros =
Now use synthetic division to see which of the zeros gives a remainder of zero.
Keep in mind that if f(x) is divided by (x – c) with a remainder of zero, then c is a zero of f(x).
Start by testing 1:
The remainder is not zero. Therefore, 1 is not a zero (or a root, or a solution) of the function.
Try 2:
The remainder is zero. Therefore, 2 is a zero (or a root, or a solution) of the function.
Thus,
x  2 is a factor of
x3  2 x 2  5 x  6 . Write the factored form and make it equal zero:


x3  2 x 2  5 x  6   x  2  x 2  4 x  3  0
Set each factor equal to zero and solve:
The solutions (the rational zeros) are  3,  1, 2.
So far, we’ve found the zeros of a polynomial whose highest power is 3. If the polynomial is of degree 4 or
higher, we should try to find more than one factor by using synthetic division.
8
Example 5
Solve x  6 x  8 x  24  0 .
4
2
Notice that the directions say “solve”. In this case, we refer to the solutions as “roots” rather than “zeros”.
70
60
50
40
30
20
10
-4
-2
2
4
6
8
-10
-20
-30
-40
9
Properties of Polynomial Equations
1. If a polynomial equation is of degree n, then counting multiple roots separately, the equation has n
roots.
2. If a + bi is a root of a polynomial equation, then the complex imaginary number a – bi is also a root
of the polynomial equation. Complex imaginary roots, if they exist, occur in conjugate pairs. This is
called the CONJUGATE ZEROS THEOREM.
Example 6
Find a polynomial function of degree 3 given that two of its roots are 3 and 2  i .
Example 7
Find all the zeros of f ( x)  x  7 x  18 x  22 x  12 given that 1  i is a zero.
Example 8
Find all the zeros of f ( x)  x  2 x  5x  4 .
4
3
3
2
2
50
45
40
35
30
25
20
15
10
5
-3
-2
-1
1
2
3
-5
-10
-15
-20
10
The next theorem says that every function defined by a polynomial of degree 1 or more has a zero, which means
that every such function can be factored.
The Fundamental Theorem of Algebra
Every function defined by a polynomial of degree 1 or more has at leas one complex zero.
The fundamental Theorem of Algebra essentially says that f(x), a polynomial function with degree n, can be
factored as:
f ( x)  a  x  c1  x  c2  x  c3 
where the n zeros are c1 , c2 , c3 ,
Example 9
 x  cn 
, cn .
Find a function f defined by a polynomial of degree 3 that has zeros 1, 2, and 4 , and
f (1)  3 . Write the function as a product of linear factors.
Example 10 Find a function f defined by a polynomial of degree 3 in which 2 is a zero of multiplicity 3,
and f (1)  4 .
Example 11 Find all the zeros of f x   x 5  x 3  2 x 2  12 x  8 . Then write f x  as a product of linear
factors.
To do example 11, first look at a computer-generated graph of f x  to see three of the five
zeros.
35
30
25
20
15
10
5
2.5
2
1.5
1
0.5
0.5
1
1.5
2
2.5
3
3.5
5
10
15
20
25
11
Example 12 Write a polynomial function of degree 4 for the graph given below.
750
700
650
600
550
500
450
400
350
300
250
200
150
100
50
8
6
4
2
2
4
6
50
100
150
200
250
300
350
12