Summary 2
Factorization of Polynomial functions
Fundamental Theorem of Algebra: Every non-constant polynomial function has a
root or zero in the complex system (the complex system includes the real numbers as a
subset)
Examples:
x = 1 is a zero of the polynomial P( x)  3x 5  2x 2  5x  4 because P(1) = 0
x = i is a zero of the polynomial P( x)  x 6  1 because P(i) = 0
Factorization Theorem: If P(x) is a polynomial of degree n>0 with leading coefficient
an then P(x) can be factored in n linear factors (not necessarily distinct) over the
complex numbers.
That is, P(x)  an (x  c1(x  c 2 ). . . (x  cn )
Note: By the Fundamental Theorem of Algebra the polynomial P(x) has a zero c 1 over
the complex numbers. Therefore x – c1 is a factor and the other factor is a polynomial
Q1(x) of degree n-1. Therefore P(x) = (x – c1) Q1(x). By the same reason if the degree of
Q1(x) is >0, then Q1(x) has a zero c2 over the complex numbers. Therefore x – c2 is a
factor and the other factor is a polynomial Q2(x) of degree n-2. Therefore P(x) = (x – c1)
Q1(x) = (x – c1) (x – c2) Q2(x). By the same reason if the degree of Q2(x) is >0, then Q2(x)
has a zero c3 over the complex numbers. Therefore x – c3 is a factor and the other factor
is a polynomial Q3(x) of degree n-2. Therefore
P(x) = (x – c1) (x – c2) Q2(x) =(x – c1) (x – c2) (x – c3) Q3(x) and so on.
Example:
Find the three zeros of the polynomial P(x)  x3  3x 2  4x  12
-3 |
1
3
4
12
-3
0 -12
1 0
4
0
Therefore x = -3 is a zero and (x+3) is a factor.
The other factor is x 2  4
x 2  4  0  x 2  4  x    4
 x  2i
Therefore, the zeros of the polynomial P( x)  x3  3x 2  4x  12 are -3, -i and i and the
factors are (x+3)(x + i)(x – i)
Conjugate Roots Theorem: Let P(x) be a polynomial with real coefficients. If the
complex number a + bi is a root then the complex conjugate a - bi is also a root.
Example:
Find the three zeros of the polynomial P( x)  4x 2  4x  37
4  ( 4) 2  4( 4)(37 ) 4   576 4  24i 1
4x  4x  37  0  x 


  3i
2( 4)
8
8
2
1
1
The zeros are the conjugate pair  3i and  3i
2
2
2
-1-
Quadratic factor Theorem: If c1 and c2 are zeroes of a polynomial function P(x) then
x 2  (c1  c 2 )x  c1c 2 is a quadratic factor of the polynomial.
Proof:
Consider the quadratic function y  ax 2  bx  c
The solutions of ax 2  bx  c  0 are c1 
 b  b 2  4ac
 b  b 2  4ac
and c 2 
2a
2a
 b  b 2  4ac  b  b 2  4ac
 c1  c 2 

2a
2a
 b  b 2  4ac  b  b 2  4ac  2b
b


2a
2a
a
  b  b 2  4ac   b  b 2  4ac  b 2  b 2  4ac
4ac c


Also : c1c 2  




2
2
2a
2a
a
4
a
4
a



b
c
c
 b
y  0  ax 2  bx  c  0  x 2  x   0  x 2     x   0
a
a
a
 a
 c1  c 2 


 x 2  c1  c 2 x  c1c 2
Example: If -5 and 8 and 2 are zeros of a polynomial P(x) then (x+5) is a linear factor
and x 2  (c1  c 2 )x  c1c 2  x 2  (8  2)x  (8)( 2)  x 2  10x  16 is a quadratic
factor.
Example: Find a polynomial a degree 4 with real coefficients and -2, 7 and 2 – 5i are
zeros.
By the conjugate root theorem if 2 – 5i is a zero then 2 + 5i is also a zero. They are
zeros of the quadratic factor
x 2  (c1  c 2 )x  c1c 2  x 2  (2  5i  2  5i )x  (2  5i )( 2  5i )  x 2  4x  20
-2 and 7 are zeros of the quadratic factor
x 2  (c1  c 2 )x  c1c 2  x 2  (2  7)x  (2)(7)  x 2  5x  14
A polynomial of degree 4 is given by


P( x )  x 2  4x  20 x 2  5x  14

 x 4  5x 3  14x 2  4x 3  20x 2  56x  20x 2  100x  280
 x 4  9x 3  26x 2  44x  280
-2-
Rational roots theorem: If c=
p
q
is a rational root (in lowest terms) of a polynomial
function f(x), then p is a factor or divisor of the constant term ao, and q is a divisor of the
leading coefficient an .
p
Pr oof : c  is rational root or zero (in lowest terms, p and q are relatively
q
prime , they have no common factors ) of
P( x )  a n x n  a n 1x n 1  . . .  a 2 x 2  a1x  a 0
n 1
n
2
p
p
p
p
 a n    a n 1  
 . . .  a 2      x  a 0  0
q
q
q
q
n 1
n2

 p   p 
p
p

   a n  
 a n 1  
 . . .  a 2    x    a 0
q
q
q
q


   
 
 

n2
  p  n 1

p
p


 p a n  
 a n 1  
 . . .  a 2    x    a 0 q
q
q
q
  

 
 


 p is a factor of a 0 because p is factor of a 0 q but not a factor of q, therefore it
has to be a factor of a 0
Also :
p
 
q
n
p

 a n  a n 1  q 
 

1
p
 . . .  a 2  
q
q
q
 a n  a n 1    . . .  a 2  
p
p
q
 a n  a n 1    .
p
 q 
 a n     a n 1 
 p 


 pa n  q  a n 1  .


 q is a factor of a n
n2
q
. .  a 2  
p
q
  
p
n2
q
. . .  a 2  
p
q
. .  a 2  
p
2n
n 1
q
  
p
n3
n3
1 n 
p
  
q
q
  
p
q
x   a 0  
p
n 1
q
  
p

x   a 0


q
x  a 0  
p
n2
n2
n
n
q
x  a 0  
p
q
x  a 0  
p
n 1 
n 1 






because q is factor of pa n but not a factor of p, therefore it
has to be a factor of a n
-3-
Example: Consider the polynomial function: P(x)  4x 3  4x 2  9x  9
p
If c  is a ze ro of P(x)  4x 3  4x 2  9x  9 , then the possible value of p are
q
 1,  3,  9 . The possible values for q are  1,  2,  4 and the possible zeros are
p
1
1
3
3
9
9
c   1,  ,  ,  3,  ,  ,  9,  , 
q
2
4
2
4
2
4
By using synthetic division you can verify that the zeros of the polynomial are
3
3
and the linear factors are (x+1), (2x + 3) and (2x – 3)
c  1,  and
2
2
Descartes' rule of signs: let P(x) be a polynomial with real coefficients and a
nonzero constant term, whose terms are arranged in order of decreasing powers of x
(with terms having a zero coefficient deleted).
a) The number of positive real roots of P(x) is either equal to the number of
variations of sign of P(x) or is less than that number by an even integer.
b) The number of negative real roots of P(x) is either equal to the number of
variations in sign of P(-x) or is less than that number by an even integer.
Example. Consider the polynomial function P(x)  9x 4  27x 3  17x 2  3x  2
b) Use Descartes’ rule of signs to complete the following table for the possible zeros of
the functions
P(x)  9x 4  27x 3  17x 2  3x  2 has three variations in sign, from +9 to -27 and -27 to
+17 and from +3 to -2. Therefore the polynomial could have three positive real solutions
or one positive real solution.
P(x)  9 x4  27 x3  17 x2  3 x  2  9x 4  27x 3  17x 2  3x  2 has one
variations in sign, from +17 to -3 . Therefore the polynomial could has one negative real
solutions
Number of positive real roots 3 or 1
Number of negative real roots 1
Number of complex roots (imaginary)
total number of roots or zeros
3
1
0
4
1
1
2
4
1 1
Use synthetic division to check that the solutions are  , , 1, 2
3 3
1 | 9
-27 17
3
-2
9 -18 -1
2
2 | 9 -18
-1
2
0
18
0
-2
9
0
-1
0
The re fore P( x )  9x 4  27 x 3  17 x 2  3x  2  ( x  1)( x  2)(9x 2  1)
P( x )  ( x  1)( x  2)(3x  1)(3x  1)
1 1
, , 1, 2
3 3
The polynomial function has one negative zero and three positive zeros.
The ze ros are 
-4-
Example 2. Consider the polynomial function P(x)  2x 4  9x 3  3x 2  16x  6
a. Give the list of all possible rational zeros of the function.
The divisors of the leading coefficient p are {1,  2,  3,  6}
The divisors of the constant term q are {1,  2}
The possible values of the rational zeros
p 
1
1
3
c    1,  2,  3,  6,  ,  ,  
q 
2
3
2
b) Use Descartes’ rule of signs to complete the following table for the possible zeros of
the functions
P(x)  2x 4  9x 3  3x 2  16x  6 has two variations in sign, from +3 to -16 and -16 to +6.
Therefore the polynomial could have two positive real solutions or no positive real
solutions.
P(x)  2 x4  9 x3  3 x2  16 x  6  2x 4  9x 3  2x 2  16x  6 has two
variations in sign, from +2 to -9 and -9 to +2. Therefore the polynomial could have two
negative real solutions or no positive real solutions.
Number of positive real roots 2 or 0
Number of negative real roots 2 or 0
Number of complex roots (imaginary)
total number of roots or zeros
c) Show that c=
1
2
2
2
2
2
0
4
1
is a zero (use synthetic division)
2
9
3  16
2
0
2
4
0
2
2
4
0
0
4
4
6
1
5
4
6
10
8
-12
0
d) Find the exact value of the other solutions : __________________
-3 |
2
10
8
-12
2
-6
4
-12
-4
12
0
2x 2  4x  4  0  x 2  2x  2  0  x 
 2  4  4(1)( 2)
2
 2  12  2  2 3

 1  3
2
2
1
The zeros or roots are  3, ,  1  3 and  1  3
2
x
The polynomials has two negative zeros and two positive zeros.
-5-
-6-