I) Polynomial Zeros
A) Rational Roots Theorem
1) Given f(x) = anxn + an–1xn–1 + + a2x2 + ax + a0, if p is a factor
of a0 and q is a factor of an, then p/q is a possible rational zero of f(x).
2) Example
List all the possible rational zeros of f(x) = 3x3 – 5x + 4
p ± (1, 2, 4)
q ± (1, 3)
p/q ± (1, 2, 4, 1/3, 2/3, 4/3)
B) Zeros
3) Zeros (by Factoring)
a) The zeros of a polynomial are the x-values of the x-intercepts
b) To find, solve f(x) = 0
c) The factored form of f(x) is f(x) = a(x – c1)(x – c2)
where c1, c2, are the zeros
d) If a + bi is a zero, a – bi is also a zero (more on this later)
e) Examples
i) Find the zeros of f(x) = x3 + 7x2 – 4x – 28
x3 + 7x2 – 4x – 28 = 0
x2(x + 7) – 4(x + 7) = 0
(x2 – 4)(x + 7) = (x + 2)(x – 2)(x + 7) = 0
x = -2, 2, -7
ii) Find a polynomial whose zeros are 3 and 2 – 3i which passes through (1, 2)
Since 2 – 3i is a zero, we know that 2 + 3i is also a zero
The factored form in general is f(x) = a(x – c1)(x – c2), so:
f(x) = a(x – 3)(x – (2 – 3i))(x – (2 + 3i))
f(x) = a(x – 3)[(x – 2)2 – (3i)2] ← We used a difference of two squares here
f(x) = a(x – 3)[x2 – 4x + 4 + 9]
f(x) = a(x – 3)[x2 – 4x + 13]
f(x) = a(x3 – 4x2 + 13x – 3x2 + 12x – 39)
f(x) = a(x3 – 7x2 + 25x – 39)
Now use the point given, (1,2)
2 = a(1 – 7 + 25 – 39)
a = -1/10
Finally, we put it all together:
= − x – 7x + 25x– 39
**Parts C and D below are tools often used in finding the zeros**
C) Long Division
1) Procedure
a) Procedure is essentially same as regular numbers
b) Divide leading terms at each step
2) Example
a) Divide x3 – 7x + 6 ÷ x – 1
x2
x – 1 x3
- x3
+x
-6
- 7x
+ x2
x2
- x2
+6
- 7x
+x
- 6x
6x
+6
-6
0
D) Synthetic Division
1) Use for polynomial division where the divisor is of the form x – c
2) Procedure
a) Add
b) Multiply
c) Carry
d) Repeat
3) By example
Find the result of x3 – 7x + 6 ÷ x – 1
1
1
1
0
1
1
-7
1
-6
6
6
0
Note what this means: = + − 6
So we have factored by one step!
To see this, multiply both sides above by x – 1 to get:
− 7 + 6 = + − 6 − 1
II) Rational Functions
A) Asymptotes
1) Vertical
a) What: A vertical line that the graph cannot cross
b) Where: Through points not in the domain
Note: if cancelation occurs, there is a hole instead of a vertical asymptote
c) Example
i) Find VA for =
x = -3
ii) Find VA for = (2x – 3)(x + 1) = 0
x = 3/2 or x = -1
2) Horizontal
a) What: A horizontal line that the graph becomes close to (and may cross at some
point) at extreme values of x
b) Where
i) Deg Num < Deg Den
2x + 1
Let f ( x) = 2
3x − 5
2(100) + 1
f (100) =
≈ 0.0067
3(100) 2 − 5
2(10000) + 1
f (10000) =
≈ 0.000067
3(10000) 2 − 5
Conclude: y = 0
ii) Deg Num = Deg Den
3x 2 + 3
Let f ( x) = 2
4x − 1
3(100) 2 + 3
f (100) =
≈ 0.75009
4(100) 2 − 1
f (10000) =
3(10000) 2 + 3
≈ 0.75
4(10000) 2 − 1
3
4
Conclude: y = ratio of coefficients of largest powers of x
iii) Deg Num > Deg Den: Asymptote is a little more complicated
⋅ The asymptote is the quotient of the division
⋅ If the degree of the numerator is only one greater than the degree of the
denominator, the quotient is a linear function and is called a "slant asmyptote"
=
3x 2 + 2 x − 4
x−2
The slant asymptote is y = 3x + 8
This is the quotient when the numerator is divided by the denominator
⋅ Let f ( x) =