1. 5 solutions/zeros Number of Solutions is... … degree of Polynomial
f(x) = x5 – 2x4 + 8x2 – 13x + 6
5 solutions/zeros

2. Determine the number of zeros by the degree (zeros can repeat).
Finding all zeros.
Same as yesterday except the final trinomial might not factor so you have to use the quadratic formula or completing the square (not calculator).
Determine the number of zeros by the degree (zeros can repeat).
Use the possible rational zeros and synthetic division until it is down to x2.
f(x) = x5 – 2x4 + 8x2 – 13x + 6 1
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Possible rational zeros:
±1,±2,±3,±6 1
__________________________ -2
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Zeros: 1,1,-2,1±i√2
x2 -2x + 3

3. Given a polynomial and one complex zero find the reaming zeros:
Practice:
x4 – 9x3 + 21x2 + 21x – 130; zero 3-2i
We know that there are 4 zeros and since one is 3 – 2i there must be a second zero that is the conjugate i
Write both zeros as roots and multiply:
(x – 3 – 2i)(x – 3 + 2i)
x2 -3x -3x + 9 – 4i2
x2 – 6x + 13
Factor and find the last 2 zeros:
x2 – 3x – 10
(x – 5)(x + 2)
x = 5, x = - 2
Are the last 2 zeros.
Divide:
x2 – 6x + 13 | x4 – 9x3 + 21x2 + 21x – 130
-( x4 – 6x3 + 13x2)
-3x3 + 8x x – 130
-( -3x x2 - 39x)
-10x2 + 60x
-(-10x2 + 60x – 130)
x2 – 3x - 10

4. Practice: How many zeros then find them all.
h(x) = x4 + 4x3 + 7x2 + 16x g(x) = x4 -2x3 – 3x2 + 2x + 2
Hint (-3 is one zero)

5. Extra Mojo Descartes’ rule of signs
The number of positive real zeros is dependent on the number of sign changes, -2 until you get to 1 or 0
The number of negative real zeros is dependent on the number of sign changes when you use f(-x), - 2 until you get to 1 or 0.
Imaginary zeros must come in pairs.
Types of Solutions is...
sign changes so
4,2,or 0 positive real solutions
f(x) = x5 – 2x4 + 8x2 – 13x + 6
f(-x) = -x5 – 2x4 + 8x2 + 13x + 6
sign change so
1 negative real solution

6. Number of Solutions and Types of Solutions is...
f(x) = 4x4 + 2x3 + 8x2 – 6x – 11
Degree 4 so 4 zeros.
sign change so
1 positive real solutions
f(-x) = 4x4 – 2x3 + 8x2 + 6x – 11
sign changes so
3 or 1 negative real solutions