2.4 Real 0’s of Polynomial Functions
Long Division
 q(x) called the quotient, r(x) called the remainder
 f(x) = dividend d(x) = divisor
 Remember when doing long division you must fill in holes
with 0, including the constant
example: if you are given x3 – 4x + 5 you will have to
rewrite it as x3 + 0x2 – 4x + 5
 Examples: f(x) = 4x3 – 8x2 + 2x - 1; d(x) = x – 1
f(x) = x3 – 1; d(x) = x + 1
Synthetic division
 Examples: f(x) = 2x3 – 3x2 – 5x – 12, d(x)=x -3
1) rewrite without the x’s, dividend: the signs stay the same,
divisor: the sign changes
3 2 -3 -5 -12
multi. on the diagonal
6 9 12
add going down
2 3
4 0
last # is always the remainder
Rewrite answer: 2x2 + 3x + 4
Examples: 2x4 – 5x3 + 7x2 - 3x -1
x -3
5x4 – 3x + 1
x-4
Remainder Theorem
 If a polynomial f(x) is divided by x – k, then the remainder is
r = f(x)
 To use the remainder theorem:
either solve the problem by long or synthetic division and
find the remainder
write your answer in the form f(x) = remainder
Example: find the remainder when f(x) = 3x2 + 7x – 20 is
divided by a) x – 2, b) x + 4
Answer:
Factor Theorem
 A polynomial of function f(x) has a factor x – k if and only if
f(k) = 0
 To do: again either use long or synthetic division
example: use the Factor theorem to determine whether the
1st polynomial is a factor of the second polynomial
Example: x -3; x3 – x2 - x - 15
Finding rational 0’s
 To do:
1) find all factors of your constant (p)
2) find all factors of your leading coefficient (q)
3) find all values of p/q – list all of them
4) pick any p/q to determine if it’s a 0 by using synthetic
division, need a remainder of 0 for it to work
5) rewrite answer & then factor it (if you get it down to x2)
6) find the 0’s by setting each factor = 0 & solving for x
Example: f(x) = 3x3 + 4x2 – 5x - 2
Upper & Lower bounds
 k is an upper bound for the real 0’s of f if f(x) is never 0 when
x>k
 k is a lower bound for the real 0’s of f if f(x) is never 0 when
x<k
 Another way to say it:
if k> 0 and every # in the last line is non-negative (positive
or 0), then k is an upper bound for the real 0’s of f.
if k <0 and the #’s in the last line are alternately nonnegative and non-positive, then k is a lower bound for all the
real 0’s of f.
Proving you have an upper or lower
bound
 Prove that the number k is an upper bound
k = 5, f(x) = 2x3 – 5x2 - 5x - 1
 Prove that the number k is a lower bound
k = -3, f(x) = x3 + 2x2 + 2x + 5
Finding real 0’s
 Use the Rational 0’s Theorem (which is what we use to find
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rational 0’s)
Find p/q (p is the factors of your constant, q is the factors of
the leading coefficient)
Pick any one you want & use synthetic division to determine
if it’s a 0 (your remainder will be 0)
Rewrite your answer: if it’s x3 or higher, then you need to
repeat what you did above. If it’s x2, you can factor it and
then solve each factor
Determine if the 0 is rational or irrational
example
 Find all real 0’s of the function, finding exact values
whenever possible. Identify each 0 as rational or irrational
f(x) = x3 + 3x2 – 3x – 9
f(x) = 3x4 – 2x3 + 3x2 + x - 2