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```Section 5.7 Notes
1. Consider the polynomial function π(π₯) = π₯ 4 + 2π₯ 3 β 7π₯ 2 β 8π₯ + 12.
The zeros of π(π₯) are:
The roots of π(π₯) are:
The factors of π(π₯) are:
The x-intercepts of π(π₯) are:
2. Solve 2π₯ 3 β 7π₯ 2 β 8π₯ + 12. = 0.
The Fundamental Theorem of Algebra states that a polynomial equation to the degree n has exactly n
roots in the set of complex numbers included repeated roots.
Therefore π₯ 3 β 5π₯ 2 + 12 has how many roots? What about 2π₯ 5 + 12π₯ 2 + 9π₯ β 22? As weβve stated
earlier in this chapter, the degree indicates the total number of roots (not necessarily the number of xintercepts).
For example, π(π₯) = π₯ 2 + π₯ + 4 has two roots because it is degree two. However, if you graph the
function it has no x-intercepts.
Descartesβ Rule of Signs:
a) The number of positive real zeros of π(π₯) is the same as the number of changes in sign of the
coefficients of the terms, or is less than this by an even number
b) The number of negative real zeros of π(π₯) is the same as the number of changes in sign of the
coefficients of the terms of π(βπ₯), or is less than this by an even number
Given the function π(π₯) = π₯ 6 + 3π₯ 5 β 4π₯ 4 β 6π₯ 3 + π₯ 2 β 8π₯ + 5, state the possible number of positive
real zeros, negative real zeros, and imaginary zeros.
a) Count the number of changes in sign for the coefficients of π(π₯)
Given π(π₯) = π₯ 6 + 3π₯ 5 β 4π₯ 4 β 6π₯ 3 + π₯ 2 β 8π₯ + 5
There are 4 sign changes, so there are 4, 2, or 0 positive real zeros
b) Count the number of changes in sign for the coefficients of
Given π(βπ₯) = (βπ₯)6 + 3(βπ₯)5 β 4(βπ₯)4 β 6(βπ₯)3 + (βπ₯)2 β 8(βπ₯) + 5
There are 2 sign changes, so there are 2, or 0 negative real zeros. Make a chart of the possible
combinations of real and imaginary zeros.
Number of Positive
Real Zeros
Number of Negative
Real Zeros
Number of Imaginary
Zeros
Total Number of Zeros
3. Find all the zeros of π(π₯) = π₯ 4 β 18π₯ 2 + 12π₯ + 80
a) First, determine the total number of zeros.
b) Then, determine the type of zeros.
c) Start to determine the real zeros using synthetic substitution.
x
x
1
0
-18
12
80
```
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