Download Section R4 B Zeros of a Function

Section R4 B Zeros of a Function
Definition: Zero (sometimes called a Root) of a Function are the values in a function's domain where
the function value equals zero.
If you visualize a function like:
the places where I cross the x-axis x = {x 1 , x2 , x3 } are the zeros.
When we work with polynomials, the number of zeros equals the highest degree of the polynomial (but
sometimes a zero will repeat). You may also see complex numbers as the zeros.
So, f x = ax 2 bxc will have 2 zeros
f x = ax 4 bx 3 cx 2 dx e will have 4 zeros
and so on...
We can find the zeros of a polynomial by factoring the original form into linear pieces like:
x2 ab xab = x a x b
Sometimes factoring does not work well, and we use formulaic constructions like:
if f x = ax 2 bxc = 0
x=
b ± b2 4ac
2a
Quadratic Formula
Sometimes we have to estimate zeros using numerical approximations, noting when f(x) changes in
value from negative to positive or positive to negative.
Let's review some of the algebraic techniques for finding roots.
Theorem: If 2 quantities a and b are multiplied together and form 0, so ab = 0, then either a = 0 or b =
0.
Example: 70=0
banana0=0
We use this idea when we factor to find zeros.
Example: f x = x2 7x 10
Note: force leading coefficient equal to 1 by dividing
Write out possible / simple arithmetic factors of the constant: 10 : {1 , 2 , 5 , 10 }
Ask yourself if you can add (or subtract) any of these together to get middle number.
In this case, yes
25 = 10 25=7
or 25 = 10 2 5 = 7
Form Factorization: f x = x 5 x 2
Find zeros:
Let f(x) = 0
x 5 x 2 = 0
x 5 = 0
x=5
x 2 = 0
x=2
Zeros of f x = {2 , 5} = x
Theorem: If p(x) is a polynomial, then:
p x = x a q x for some polynomial q(x), if and only if x = a is a zero of p(x) (i.e.:
p a = 0 )
Example: Sow that x = 3 is a zero of f x = 6x 3 19 x 2 x6
If x = 3, then f (3) = 0. So, test:
f 3 = 6 33 19 32 3 6
Yeah!
= 6 2719 93 6 = 0
Now, we can use this known zero to help us find other zeros:
Right now we know that (x-3) is a linear factor of f(x). So f x = x 3 q x . We just need to
figure out q(x).
Option 1: Long Division:
6x 2 x 2
x 3
6x 3 19x2 x 6
6x 3 18 x 2 x2 x 6
x 2 3x 2x 6
2x 6 0
So: f x = x 3 6x 2 x2
6x 2 x2 = 0
3x 2 2x1 = 0
3x 2 = 0
2x1 = 0
2
1
x=
x =
3
2
x 3 = 0
x= 0
{
Zeros of f x = x = 1 2
,
,3
2 3
}
Option 2: Synthetic Division
p
(fully reduced) is
q
a zero (when said polynomial has integer coefficients) then p is a factor of the constant
term and q is a factor of the leading coefficient.
Rational Zero Theorem: if f x = polynomial and the rational number
Synthetic Review:
f x =
6x 3
19x2
3
6
-19
3 6
6
-1
Known Zero x
6
Coefficient of each
3 1 3 2 term in decreasing order
1
6
-2
0
Remainder
Note: We want the remainder to be zero.
This technique is particularly helpful for high degree polynomials.
Yielding: f x = x 3 6x 2 x2
Now we manage q x = 6x 2 x 2
From the leading coefficient 6 = {1 , 2 , 3 , 6} “q”
constant 2 = {1 , 2 } “p”
Possible Roots: ±
1
1
1
2
2
2
2
1
,± ,± ,± ,± ,± ,± ,±
1
2
3
6
1
2
3
6
Test:
1/1 = 1
6
-1
-2
16
1 5
6
5
3
Note: the remainder is NOT zero, 1 is not a factor.
1/ 2
Aha!
6
-1
-2
1
6
2
1
4
2
6
-4
0
q x = x 1
6x 4 2
6x 4 = 0
2
x =
3
1
x =
2
as before
Option 3: Quadratic Formula
We realize: f x = x 3 6x 2 x2
We assume
6x 2 x 2 = 0
ax 2 bxc = 0
So: a = 6, b = -1, c = -2 and apply form:
x=
1± 12 4 62
b ± b2 4ac
1± 148
1±17
=
=
=
2a
26
12
12
17
8
2
17
6
1
=
=
;
=
=
12
12
3
12
12
2
{
Zeros of f x = x = }
1 2
,
, 3 as before.
2 3
NOTE: This is the most robust method for 2nd degree polynomials but other preceding methods work
for any degree.