I) Polynomial Zeros

A) Rational Roots Theorem 1) Given f(x) = anx

n + an–1xn–1 + � + a2x

2 + 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 - 6

x – 1 x3 - 7x + 6 - x3 + x2

x2 - 7x - x2 + x

- 6x + 6 6x - 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 0 -7 6

1 1 6

1 1 -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

Let 53

12)(

2−

+=

x

xxf

0067.05)100(3

1)100(2)100(

2≈

−

+=f

000067.05)10000(3

1)10000(2)10000(

2≈

−

+=f

Conclude: y = 0 ii) Deg Num = Deg Den

Let 14

33)(

2

2

−

+=x

xxf

75009.01)100(4

3)100(3)100(

2

2

≈

−

+=f

75.01)10000(4

3)10000(3)10000(

2

2

≈

−

+=f

4

3=

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"

⋅ Let 2

423)(

2

−

−+=

x

xxxf

The slant asymptote is y = 3x + 8 This is the quotient when the numerator is divided by the denominator