College Algebra

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Previous Lessons
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:

- Solve a Polynomial Inequality
- Solve a Rational Inequality


Steps for Solving Polynomial Inequalities

Start by bringing everything to one side. And then replace the inequality symbol with an equal sign, so you get the form `P(x) = 0`.

Now solve for the zeros of `P(x)`.

These x-values split the x-axis into intervals. Choose a test value from each interval.

Plug each test value into P(x) and determine whether the output is negative or positive for each interval.

Then identify your solution set.



`x^5+x^2 >= 5x^3+5`

Now let's go ahead and subtract `5x^3` and subtract `5` on both sides. So that we get `x^5-5x^3+x^2-5 >= 0`. So the related equation is `x^5-5x^3+x^2-5 = 0`. Now we need to solve for the zeros of this polynomial. So let's use the Rational Zeros Theorem. So P can include `+-1," and "+-5`, and q can include `+-1`. So then `p/q` can include +1, -1, +5, and -5.

So let's let `P(x) = x^5+0x^4-5x^3+x^2+0x-5`. Then if we divide `(P(x))/(x-(-1))`, we have 1 in the corner, 1, 0, -5, 1, 0, -5. So bring down the 1, multiply by 1, add to zero, multiply by 1, add to -5, multiply by 1, add to 1, multiply by 1, add to zero, multiply by 1, add to -5. And get -8, which is our remainder. That means by the Remainder Theorem, `P(1) = -8`, and 1 is not a zero of our function P(x).

Now let's go ahead and check -1. So put -1 in the corner, bring down the 1, multiply by -1, add to zero, multiply by -1, add to -5, multiply by -1, add to 1, multiply by -1, add to zero, multiply by -1, add to -5. So the remainder of `P(x)/(x-(-1)) = 0`, and by the Remainder Theorem, `P(-1) = 0`. So -1 is a zero of P(x), and that means that `(x-(-1))` or `(x+1)` is a factor of P(x). And here are the coefficients of our other factor. It's going to be one degree less than 5. So we have `(x^4-x3-4x^2+5x-5)`.



So we have `P(x) = (x+1)(x^4-x^3-4x^2+5x-5)`. And this second part, we can just go ahead and factor by grouping. So in our first group, let's write `(x^4-4x^2-5)`, and in our second group `(-x^3+5x)`. So then we have `(x+1)`, `x^2*x^2` makes `x^4`, and -5 and positive 1 multiply to make -5, but add to make -4. And now in the last piece, we can pull out a -x, which gives us `x^2-5`.

So we can factor out an `x^2-5`. So we have `x^2+1-x`. Now if we set everything equal to zero, we can solve for the zeros of P(x). And by the Principle of Zero Products, we have `x = -1` or `x = +- sqrt(5)`. And then we can use the quadratic formula to factor the last piece. `a = 1`, b is equal to the coefficient of x, which is -1, and c is equal to 1. So that `x = (-b+-sqrt(b^2-4ac))/(2a)`. So this gives us `x = 1/2+-sqrt(3)/(2i)`. Since this is an imaginary zero, we're going to throw it out. We only want the rational zeros, which are the -1 and the `+-sqrt(5)`. Our rational zeros divide our x-axis into 4 intervals. So that we have the interval, `(-oo, -sqrt(5)), (-sqrt(5), -1), (-1, +sqrt(5))," and "(+sqrt(5), oo)`.

If we choose a test value from each interval, we can choose -3, -2, 2, and 3. And then let's plug each of these test values back into our function to determine whether our function is positive or negative in each of these intervals.

So `P(-3) = -104`, `P(-2) = 7`, `P(2) = -9`, and `P(3) = 112`. That's a negative, a positive, a negative, and a positive. And since our inequality is greater than or equal to zero, we're only interested in our positive intervals. That's the `(-sqrt(5), -1)`, and `(sqrt(5), oo)`. Since our inequality is greater than or equal to, we include the endpoints, so we're going to use brackets in each of our intervals. So our solution set is `[-sqrt(5), -1] cup [sqrt(5), oo)`.



Steps for Solving Rational Inequalities

Bring everything over to one side. Replace the inequality symbol with an equal sign, so you get the form `P(x) = 0`.

Next solve for critical values of P(x). These include all values of x for which P(x) is either undefined or zero.

These x-values split the x-axis into intervals. Choose a test value from each interval.

Plug each test value into P(x) and determine whether the output is negative or positive for each interval.

And then identify your solution set.



`(x+4)/(x+5)-(x+2)/(x-1) <= 0`.

So let's start by setting everything equal to zero. And we'll call everything on the left side, P(x). A look at the denominator shows us that P(x) is not defined for `x = -5` and `x = 1`. So we've already got two critical values; they're -5, 1.

Now, solve for x. Multiply both sides by `(x+5)(x-1)`, our LCD. So that we get `(x+4)(x-1)-(x+2)(x+5) = 0`. `x^2-x+4x-4`, and the second part, we get `[x^2+5x+2x+10]`.

So we have `x^2-x+4x-4-x^2-5x-2x-10 = 0`. So our `x^2` terms cancel, and we can combine the -x, the 4x, the -5x, and the -2x. That gives us -4x, and -4 and -10 become -14. So, add 14 to both sides, and we get `-4x = 14` or `x = 14/-4`, which is the same as `-7/2`. So we have one more critical value at `-7/2`.



The critical values divide our x-axis into four intervals. So we have `(-oo, -5), (-5, -7/2), (-7/2, 1)," and "(1, oo)`. Now let's choose test values from each interval to determine the sign of our function in each interval. So, -6, -4, 0, and 2 are my choices.

And if we plug them into our function, we get about 1.43, `-0.4`, 2.8, and about -3.14. That's a positive value, a negative value, a positive value, and a negative value.

And since our inequality is less than or equal to, we're only interested in our negative intervals. That's `(-5, -7/2)` and `(1, oo)`. And because of the equals sign, we want a bracket anywhere that `-7/2` is. Not the -5 and 1, because our function is undefined there. Our solution set is `(-5, -7/2] cup (1, oo)`.