LESSON 15 (2.3) REAL ZEROS OF POLYNOMIAL FUNCTIONS
You should learn to:
1. Use long division to divide polynomials by other polynomials.
2. Use long division, factoring, and the Quadratic Formula to find real zeros of polynomial functions.
Terms to know: long division (of polynomials), remainder
Knowing how to divide one polynomial into another can help you find the zeros of polynomial functions. This
process is known as long division of polynomials. It is very similar, as a process, to long division of whole numbers.
Example 1: Divide 465 by 11,
write the remainder as a fraction.
Example 2: Divide 3 x 2  x  2 by x  1 ,
write the remainder as a fraction.
3x  4 
42 113
11 465
42
3
11
2
x 1
x  1 3x 2  x  2
(44)
3x  4  x21
 (3 x 2  3 x)
25
4x  2
22
(4 x  4)
3
2
Example 3:
3
2
a. Divide x  2 x  9 by x  3 .
x2  x  3
x  3 x3  2 x 2  0 x  9
x2  x  3
 ( x3  3x 2 )
x2  0x
( x 2  3 x )
3x  9
(3 x  9)
0
The long division process produced a quotient with no remainder. This means that x  3 is a factor of
x3  2 x 2  9, so x  3 is a zero for the function f  x   x3  2x 2  9 .
b. Find all other zeros.
x2  x  3  0
x
1  12  4(1)(3)
2(1)
1  1  12
2
1  11
x
no more real zeros
2
x
The degree of the polynomial determines the number of zeros. Zeros can be real or imaginary. They can also be
repeated (remember multiplicity).
Example 4:
a. Factor x 4  2 x 2  3 .
b. Divide x 4  2 x 2  3 by x 2  1 .
x2  3
x4  2 x2  3
x2  1 x4  2x2  3
( x 2  3)( x 2  1)
 ( x4  x2 )
 3x 2  3
c. Find the real zeros of g  x   x4  2 x2  3 .
(3x3  3)
0
( x 2  3)( x 2  1)  0
x2  3  0
x2  1  0
x2  3
x 2  1
x2  3
x 2  1
x 3
two real
x   1
non-real zeros
zeros
d. How many non-real (imaginary) zeros do you think there are for g  x  ? Two
Example 5:
a. Use long division and factoring to write f  x   6 x3  19 x2  16 x  4 in factored form, given that x  2 is a
factor of 6 x3  19 x 2  16 x  4 .
6 x2  7 x  2
x  2 6 x 3  19 x 2  16 x  4
f  x   6 x3  19 x 2  16 x  4
f ( x)  ( x  2)(6 x 2  7 x  2)
f ( x)  ( x  2)(2 x  1)(3x  2)
 (6 x 3  12 x 2 )
 7 x 2  16 x
(7 x 2  14 x)
2x  4
 (2 x  4)
0
factored form: f  x   ( x  2)(3x  2)(2 x 1)
b. Find zeros. x  2 (crosses) x 
2
3
(crosses) x  12 (crosses)
3
2
c. Sketch the graph of y  6 x  19 x  16 x  4 without using a calculator.
E.B.:

Sometimes, your calculator can enable you to find one or more zeros, allowing you to use long division, factoring, or
the quadratic formula to find the others. You may have to use repeated long divisions.
If so, always divide into the quotient - rather than the original function.
Example 6: Find the exact values of the zeros for h  x   x4  3x3  2x2  x  1 . Hint: Use your
calculator to get started.
x  1 is a zero so ( x  1) is a factor
x3  2 x 2  1
x  1 x 4  3x3  2 x 2  x  1
( x 4  x 3 )
2 x3  2 x 2
(2 x3  2 x 2 )
 x 1
 ( x  1)
0
But notice that at x  1 it bounces which means that x  1 has multiplicity 2
so ( x  1) is a factor twice.
x2  x 1
h  x   x 4  3x3  2 x 2  x  1
x  1 x3  2 x 2  0 x  1
h( x)  ( x  1) 2 ( x 2  x  1)
( x 3  x 2 )
x2  0 x
zeros :
x  1 (multiplicity 2)
( x 2  x )
and
 x 1
( x  1)
0
ASSIGNMENT 15
x
1  12  4(1)(1)
2(1)
x
1  1  4
2
x
1  5
2
Pages 127-130 (1, 11, 43 (use long division to get started), 53, 63, 90 (ignore the
instructions - just find the exact values of the zeros. See Example 6), 97, 107, 110)
Page 113: 19, 21, (show end behavior using directed arrows), 35, 40, 44 (factor first)