### Assignments:

Unfinished Assignment Study Questions for Lesson 23

### 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.

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).