4.3.1 Real Zeros of Polynomial Functions - Part I
Integer division:
quotient remainder dividend divisor
1
dividend 11
 5

11 = (2)(5) + 1
2
divisor 2
quotient
remainder
dividend = divisor  quotient + remainder
Polynomial division:
Quotient: q(x)
Example:
x
+3
Dividend: f(x)
x-1
Divisor: d(x)
2
x
x2
+2x
-x
3x
3x
-1
-1
-3
2
Remainder: r(x)
Note: you repeat the process until:
degree(remainder) < degree(divisor)
So we have:
2
x 2  2x - 1
= (x + 3) +
x -1
x 1
Or (mult. by x - 1): x2 + 2x - 1 = (x - 1)(x + 3) + 2
Note: dividend = divisor  quotient + remainder
or, f(x) = d(x)q(x) + r(x), where degree r(x) < degree d(x)
This is known as the “Division Algorithm”.
4.3.1-1
Synthetic division
 useful for (quick) division by x - a (a is a constant)
 does not work for other cases
x 2 + 2x - 1
Example:
here, a = 1
x - 1
Write out
a (coefficients of numerator go here):
1
1
2
-1
__________________
1
 (this is "brought down" from above)
Scheme is "multiply and add":
1
1
1
2
1
3
coefficients
of quotient
-1
3
2
remainder
degree(quotient) = degree(dividend) – 1 = 1 (this example)
so quotient = x + 3, remainder = 2
Result:
2
x 2 + 2x - 1
= (x + 3) +
x - 1
x - 1
Notes:
 can use synthetic div. to divide by x + 1: a = -1 ]
 if there are terms missing in dividend, have to supply 0:
3x3 - 2x - 1
would be written as:
3 0 -2 -1
4.3.1-2
Remainder and Factor Theorems
 divide a polynomial f(x) by (x - c), c a constant
 by the Division Principle we have:
f(x) = (x - c)q(x) + d
 where d is a constant (= 0 if division comes out “even”)
So:
f(c) = (c - c)q(c) + d = d
REMAINDER THEOREM
If you divide a polynomial f(x) by (x - c) and obtain
remainder d, then:
f(c) = d
Application:
Given f(x) = 3x2 - 15x - 5, find f(8)
Way 1: (Brute force) f(8) = (3)(82) - (15)(8) - 5 =
Way 2: (Using Remainder Theorem)
Just find remainder of division (3x2 - 15x - 5)  (x - 8) :
8
3
-15
-5
FACTOR THEOREM
1. if c is a zero of f(x) then (x - c) is a factor of f(x)
2. if (x - c) is a factor of f(x) then c is a zero of f(x)
Example:
if 5 is zero of a polynomial, then (x -5) is a factor
if (x - 5) is a factor of a polynomial, then 5 is a zero
4.3.1-3
Solutions, zeros, and factors
 think about the polynomial x2 + 2x - 3
 it can also be written (x + 3)(x - 1)
 so (x - 1) is a factor of the polynomial
 and, clearly, 1 is a zero of the polynomial x2 + 2x - 3
 think about the equation x2 + 2x - 3 = 0
 solve: (x + 3)(x - 1) = 0  x = 1, -3
 one solution of the equation is x = 1
Relationship between zeros, factors, and solutions
The following are equivalent:




c is a solution of the equation f(x) = 0
c is a zero of the polynomial f
x - c is a factor of the polynomial f
x – c divides f(x) with no remainder (evenly)
Division and factoring




if I tell you that one factor of 24 is 6
can you find the other factor?
of course - do 246 = 4 !!!
so 24 = 6*4
 same holds for polynomials!
 if x + 4 is a factor of x3 - 28x - 48
 (we know it is, because -4 is a zero -- try it!)
 then do (by synthetic division)
(x3 - 28x - 48) (x + 4) = x2 - 4x - 12
 and so (x3 - 28x - 48) = (x + 4)(x2 - 4x -12)
4.3.1-4
Applications
Example 1:
Construct a polynomial P(x) whose zeros are 3, -1 and 7.
Using Factor Theorem (Part 1), we can write
P(x) = (x - 3)(x + 1)(x - 7)
Example 2:
Given that x = -4 is a solution of f(x) = x3 - 28x - 48 = 0,
factor f(x) completely.
x = -4 is a solution of the equation, so -4 is a zero of f(x) ,
so (x - (-4)) = (x + 4) is a factor of f(x).
What is the other factor, q(x)? Find it by division,
(x3 - 28x - 48) (x + 4):
-4
1
0
-28
-48
q(x) = x2 - 4x - 12
so f(x) = (x + 4)(x2 - 4x -12) = (x + 4)(x - 6)(x + 2)
Voila! Factored completely!
4.3.1-5