Download A polynomial of degree n may be written in a standard form:

POLYNOMIALS AND THEIR ZEROS
A polynomial of degree n may always be written in
f ( x)  a n x n  a n1 x n1  ... a1 x  a0
 a standard form:
where a n is the leading coefficient and (0, a 0 ) is the y-intercept, or
 a complete factored form:
f ( x)  an ( x  xn )( x  xn 1)  ... ( x  x1)
where a n is the leading coefficient and the numbers x1, …, xn are zeros of the polynomial
f, which means that:
f ( xn )  0, f ( xn 1)  0,..., f ( x1)  0
In general, the zeros may be complex numbers.
This is at the heart of The Fundamental Theorem of Algebra whose consequence
is that a polynomial of degree n has exactly n complex zeros, where complex numbers
include real numbers.
Note:
 If a number z is a real zero of a function f, then a point (z, 0) is an x-intercept of the
graph of f.
 The non-real zeros of a function f will not be visible on a xy-graph of the function.
Examples:
Standard
Form
f ( x)  3 x 2  3 x  6
Factored
Form
f ( x)  3( x  1)( x  2) h( x)  2( x  1)( x  1) 3
g ( x)  2( x  1) 2 ( x  i)( x  i)
x1  1, x 2  2
x1  1, x2  1,
x1  1, x2  1,
x3  1, x4  1
x3  i, x4  i
Zeros
h( x)  2 x 4  4 x 3  4 x  2 g ( x)  2 x 4  4 x 3  4 x 2  4 x  2
FINDING ZEROS OF POLYNOMIALS
If f ( x)  an ( x  xn )( x  xn 1)  ... ( x  x1) then the zeros are shown explicitly ( x1 ,...,x n )
but if f is not given in a complete factored form then depending on the degree different
techniques apply.
Examples
For a polynomial of degree 2, a quadratic function, we can always use the Quadratic
Formula to find the zeros. In some cases, factoring is possible instead.
2
1. Let f ( x)  3x  3x  6 . Find the zeros of f, i.e. solve f(x) = 0
Factoring
Quadratic Formula
3x 2  3x  6  0
a3
3( x  x  2)  0
3( x  2)( x  1)  0
x  2  0 x 1  0
x  2
x 1
2
OR
b3
c  6
 b  b 2  4ac
x
2a
x
 3  (3) 2  4(3)( 6)
2(3)
 3  81  3  9

6
6
39
39
x1 
2
x2 
1
6
6
x
2
2. Let f ( x)  x  8 . Find the zeros of f, i.e. solve f(x) = 0.
Factoring
x2  8  0
Quadratic Formula
a 1
( x  i 8 )( x  i 8 )  0
x  i 8  i2 2
or
x  i 2 2
x
OR
b0
c 8
 b  b 2  4ac
2a
0  (0) 2  4(1)(8)  32  4 2
x


2(1)
2
2
x1  i 2 2
x2  i 2 2
FINDING ZEROS OF POLYNOMIALS
For a polynomial of degree n > 2 we can try factoring techniques. If they do not apply
easily or at all, there are theorems that help in narrowing down the candidates for zeros. To
check if a particular number, x1, indeed is a zero of a polynomial we can divide the
polynomial by the factor (x – x1). If the remainder is equal to zero than we can rewrite the
polynomial in a factored form as ( x  x1 )  f1 ( x) where f1 ( x) is a polynomial of degree
n  1 . This process can be continued until all zeros are found.
Factoring
f ( x)  2 x 4  4 x 3  4 x  2
f ( x)  2( x  1)( x  1) 3  0
x1  1 x2  1 x3  1 x4  1
Division by linear factors of the form x - c
Is x = 1 a zero of f? Use synthetic division to check if (x – 1) divides f without a remainder:
1 -2
-4 0 4 2
-2 -6 -6 -2
____________________
-2
-6
-6 -2
0
Since the remainder = 0 then the polynomial f can be rewritten as
f ( x)  ( x  1) f1 ( x)  ( x  1)( 2 x 3  6 x 2  6 x  2)
Is x = 1 a zero of f1?
1 -2 -6 -6 -2
-2 -8 -14
____________________
-2 -8 -14 -16
Since the remainder is not 0 then the polynomial f has only one zero x = 1.
Is x = -1 a zero of f ?
-1 -2 -6 -6 -2
2
4
2
____________________
-2 -4
-2
0
Since the remainder = 0 then the polynomial f can be rewritten as
f ( x)  ( x  1)( x  1) f 2 ( x)  ( x  1)( x  1)(2 x 2  4 x  2)
To find the remaining two zeros we can always use the Quadratic Formula.