College Algebra

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Open Chapter Ch. 1: Graphs, Functions, and Models
Lesson #1 Graphing
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Lesson #2 Graphing Functions
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Lesson #3 Slope of Linear Functions
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Lesson #4 Linear Equations and Modeling
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Lesson #5 Linear Equations, Functions, Zeros and Applications
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Lesson #6 Linear Inequalities
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Open Chapter Ch. 2: More on Functions
Lesson #7 Analyzing Functions
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Exam Exam 1
Lesson #8 Algebra for Functions
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Lesson #9 The Composition of Functions
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Lesson #10 Symmetry
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Lesson #11 Transformations
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Lesson #12 Equations of Variation
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Open Chapter Ch. 3: Quadratic Functions and Equations; Inequalities
Lesson #13 Complex Numbers
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Lesson #14 Quadratic Equations, Functions, Zeros and Models
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Lesson #15 Analyzing Graphs of Quadratic Functions
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Lesson #16 Rational and Radical Equations
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Exam Midterm Exam
Lesson #17 Absolute Value Equations and Inequalities
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Open Chapter Ch. 4: Polynomial Functions and Rational Functions
Lesson #18 Polynomial Functions and Models
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Lesson #19 Graphing Polynomial Functions
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Lesson #20 Polynomial Division
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Lesson #21 The Zeros of Polynomial Functions
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Lesson #22 Rational Functions
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Lesson #23 Polynomial and Rational Inequalities
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Open Chapter Ch. 5: Exponential Functions and Logarithmic Functions
Lesson #24 Inverse Functions
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Lesson #25 Exponential Functions and Graphs
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Lesson #26 Logarithmic Functions and Graphs
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Exam Exam 3
Lesson #27 Properties of Logarithmic Functions
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Lesson #28 Solving Exponential Equations and Logarithmic Equations
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Lesson #29 Applications and Models: Growth and Decay; Compound Interest
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Open Chapter Ch. 6: Systems of Equations
Lesson #30 Systems of Equations in Two Variables
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Lesson #31 Systems of Inequalities and Linear Programming
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Lesson #32 Nonlinear Systems of Equations and Inequalities
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Lesson #33 Sequences and Series
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Exam Final Exam

Assignments:

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Lesson Objectives:

- The Fundamental Theorem of Algebra
- Non Real and Irrational Zeros
- The Rational Zeros Theorem
- Descartes' Rule of Signs


The fundamental theorem of algebra states that every polynomial function which has a degree `n >= 1`, has at least one zero in the set of complex numbers.

And note that polynomials may have coefficients that are complex numbers.

The zeros of a polynomial function f(x) are the solutions of `f(x) = 0`.



The following are zeroes of a polynomial function of degree 3.

-4, 3i, and -3i.

Find the polynomial function. A polynomial function of degree 3 can have at most 3 zeros, and the leading term will have `x^3`.

So we can find our factors given our zeros. They're `(x-(-4)), (x-(3i))," and "(x-(-3i))`, or `(x+4)(x-3i)(x+3i)`. So f(x) is equal to some constant `a_n`, that can't equal zero, times `(x+4)(x-3i)(x+3i)`.

So let's let `a_n = 1`, to get the simplest possible polynomial function. And now let's multiply our factors. We have `(x+4)(x^2-(3i)^2)`. And remember that you can quickly multiply something of the form `(a+b)(a-b)`, because it's equal to `("first term")^2-("second term")^2`.

And now, before we FOIL, let's take care of the `(3i)^2`. That's `9i^2`, where `i^2 = -1`. So we have `-9(-1)` or positive 9. Now if we FOIL, we get `x^3+9x+4x^2+36`.

So our polynomial function is `f(x) = x^3+4x^2+9x+36`.



Non Real Zeros

Non real zeros occur in conjugate pairs. This means that a polynomial function with real coefficients has a complex zero `a + bi`, where `b != 0`, only if its conjugate `a-bi` is also a zero.

Irrational Zeros

Irrational zeros occur in conjugate pairs, as well. This means that a polynomial function with rational coefficients has an irrational zero of the form, `a+csqrt(b)`, only if its conjugate, `a-csqrt(b)`, is also a zero.

a, b, and c are rational and b is not a perfect square.



Given the following zeros for a polynomial function of degree 4:

-i and `2-sqrt(7)`

What are the other zeros? So a polynomial function of degree 4 can have at most four zeros. And since non real zeros occur in conjugate pairs, we know that if -i is a zero, positive i must also be a zero. And since irrational zeros occur in conjugate pairs as well, since `2-sqrt(7)` is a zero, `2+sqrt(7)` must also be a zero.

So the zeros are -i and positive i, and `2-sqrt(7)` and `2+sqrt(7)`.


The Rational Zeros Theorem.

If you have a polynomial function of the form `a_nx^n+a_(n-1)x^(n-1)` plus etc..

Given all integer coefficients, let `p/q` be a rational number, where p and q don't have any common factors other than -1 and 1.

Then `p/q` is a zero of P(x), if p is a factor of `a_0` and q is a factor of `a_n`.



Given `f(x) = x^3+4x^2-3x-12`, find the rational zeros, and all other zeros. And then factor f(x) into linear factors.

According to our Rational Zeros Theorem, if we let p be a factor of -12 and q be a factor of 1, then `p/q` is a zero of P(x). So possibilities for p include `+-1`, `+-2`, `+-3`, `+-4`, `+-6`, and `+-12`. And then possibilities for q are `+-1`, so `p/q` will include 1, -1, 2, -2, 3, -3, 4, -4, and 6, and -6, and 12, and -12. So these are all possible zeros of our function.



We can either plug each of these values back into our function, and see if we get zero, to determine if it's a zero or not, or we can do a much quicker method, by using synthetic division.

The Remainder Theorem states that if `(f(x))/(x-1)` has a remainder of 0, then `f(1) = 0`. So let's put 1 in the corner, and then 1, 4, -3, and -12. Then bring 1 down, multiply by 1, add to 4, multiply by 1, add to -3, multiply by 1, add to -12, and get -10.

So by the Remainder Theorem, f(1) = -10. 1 is not a zero for our function.

Now let's try -1. Bring down the 1, multiply by -1, add to 4, multiply by -1, add to -3, multiply by -1, add to -12, and `f(-1) = -6`, which means that -1 is not a zero of our function.



So 1 and -1 are not zeros.

Let's try 2. Bring down the 1, multiply by 2, add to 4, multiply by 2, add to -3, multiply by 2, add to -12, and the remainder is 6. So by the Remainder Theorem, `f(2) = 6`. That means that 2 is not a zero.

Let's try -2. Bring down the 1, multiply by -2, add to 4, multiply by -2, add to -3, multiply by -2, add to -12, and you get a remainder of 2. That means that `f(-2) = 2` by the Remainder Theorem. That means that -2 is not a zero of our function.



Now let's check 3. Bring down the 1, multiply by 3, add to 4, multiply by 3, add to -3, multiply by 3, add to -12, and by the remainder theorem, `f(3) = 30`. 3 is not a zero of our function.

Let's check -3. Bring down the 1, multiply by -3, add to 4, multiply by -3, add to -3, multiply by -3, add to -12. That means that `f(-3) = 6`. -3 is not a zero of our function.



Now let's check 4. Bring down the 1, multiply by 4, add to 4, multiply by 4, add to -3, multiply by 4, add to -12, and we get 104. And so by the Remainder Theorem, `f(4) = 104`. So 4 is not a zero of our function.

Let's try -4. Bring down the 1, multiply by -4, add to 4, multiply by -4, add to -3, multiply by -4, and add to -12, and we get zero. And so by the Remainder Theorem, `f(-4) = 0`, which means -4 is a zero of our function.



So our rational zero is -4. This means that `x-(-4)` is a factor of our function, the same as `x+4`. These are the coefficients of our quotient, and it's going to be one less than the power of our function, so we have `x^2+0x-3`, or in other words, `x^2-3`. So if we set all of this equal to zero, we can find all of our zeros.

So using the Principle of Zero Products, we have `x+4 = 0" or "x^2-3 = 0`. So `x^2 = 3`, or `x = +-sqrt(3)`, and we already have that `x = -4`. So our other zeros are `+sqrt(3)` and `-sqrt(3)`.

And now we can factor f(x) into linear factors. We have `f(x) = (x+4)(x+sqrt(3))(x-sqrt(3))`.



Descartes' Rule of Signs

If a polynomial function, P(x), is written in ascending or descending order, and has real coefficients and a nonzero constant term, then:

The number of positive real zeros is either equal to the number of sign variations in P(x) or less than the number of sign variations in P(x) by a positive even integer.

The number of negative real zeros is either equal to the number of sign variations in P(-x) or less than the number of sign variations in P(-x) by a positive even integer.



Use Descartes' Rule of Signs to determine the number of positive and negative real zeros in f(x).

We're given `f(x) = 5x^5-2x^2+x-1`. So we can use Descartes' Rule of Signs, because our function is written in descending order, has real coefficients, and a non-zero constant term.

Now let's count our sign variations. Our first term is positive and our second term is negative, so there is one sign variation. Our second term is negative, our third term is positive, so we have another sign variation. And then our third term is positive, and our last term is negative, so we have another sign variation. That's 3 sign variations. So this means that the number of positive real zeros is either 3 or less than 3 by a positive even integer. So the number of positive real zeros is either 3 or 1.

Now let's check how many negative real zeros there are by plugging in -x. So we have `5(-x)^5-2(-x)^2+(-x)-1`. So this is the same as `-5x^5-2x^2-x-1`, and since all of our terms are negative, there are zero sign variations. There are no negative real zeros.