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Solve each polynomial inequality in Exercises $1-42$ and graph the solution set on a real number line. Express each solution set in interval notation.$$2 x^{2}+x<15$$

The solution set is $\left(-3, \frac{5}{2}\right)$

Algebra

Chapter 2

Polynomial and Rational Functions

Section 7

Polynomial and Rational Inequalities

Exponents and Polynomials

Rational Functions

McMaster University

Harvey Mudd College

Baylor University

University of Michigan - Ann Arbor

Lectures

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Solve each polynomial ineq…

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okay for this question were given an inequality. First thing we're gonna want to do to it is solvent. So that zero is on one side of our equation, like so from here, we're gonna try and factor the polynomial. So we need to numbers that multiply together to give us negative 15. And when we add them, uh, with the multiplication of this to, we're gonna get a positive one. So it seems like plus three and minus five would do that minus five. X plus x x. Keep this deposit X minus five times faster. Three gives us to make 15. Great. So our points of interest here are gonna be the answers to these, which are the roots of our polynomial. And so for this one, it's just going to be negative three. And for this one, it's going to be positive. 5/2 over to. Okay, so then we want to find the intervals, which this polynomial is less than zero. So let's start by figuring out whether each interval here is positive or negative. If we plug a number that's less than negative three into these equations, we're going to get so say we do negative for We're gonna get negative 13 for this one and the negative one for this one. The big takeaways that we have to negative numbers. And when we multiply them together, we get a positive number. If we plug something in between these two numbers zero in, we're gonna get negative. Five Positive. Three again. The big take away, one negative one Positive. That means this is gonna be negative. And if we do it with this interval, we could plug in, for example. Three. We're gonna get a positive one in a positive six to positive numbers multiplied. Gives us a positive. So what this graphed out could look like is something like this, Right? A problem. You could look like this. And if we're interested in where it's less than zero, that's just gonna be this middle interval, which we can write as negative Three up to 5/2. Making sure to use the round brackets because we don't actually want to include those exact points

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