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Solving Inequalities: An Overview (page 3 of 3) Sections: Linear inequalities, Quadratic inequalities, Other inequalities General Polynomial Inequalities
First, I factor to find the zeroes: x5 + 3x4 – 23x3 – 51x2 + 94x + 120 = (x + 5)(x + 3)(x + 1)(x – 2)(x – 4) = 0 ...so x = –5, –3, –1, 2, and 4 are the zeroes of this polynomial. (Review how to solve polynomials, if you're not sure how to get this solution.) To solve by the Test-Point Method, I would pick a sample point in each interval, the intervals being (negative infinity, –5), (–5, –3), (–3, –1), (–1, 2), (2, 4), and (4, positive infinity). As you can see, if your polynomial or rational function has many factors, the Test-Point Method can become quite time-consuming. To solve by the Factor Method, I would solve each factor for its positivity: x + 5 > 0 for x > –5; x + 3 > 0 for x > –3; x + 1 > 0 for x > –1; x – 2 > 0 for x > 2; and x – 4 > 0 for x > 4. Then I draw the grid:
...and fill it in:
...and solve:
Then the solution (remembering to include the endpoints, because this is an "or equal to" inequality) is the set of x-values in the intervals [–5, –3], [–1, 2], and [4, positive infinity]. As you can see, if your polynomial or rational function has many factors, the Factor Method can be much faster. Rational Inequalities
First off, I have to remember that I can't begin solving until I have the inequality in "= 0" format.
Now I need to convert to a common denominator:
...and then I can simplify:
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Elizabeth Stapel 1999-2011 All Rights Reserved
The two factors are –x + 6 and x – 3. Note that x cannot equal 3, or else I would be dividing by zero, which is not allowed. The first factor, –x + 6, equals zero when x = 6. The other factor, x – 3, equals zero when x = 3. Now, x cannot actually equal 3, so this endpoint will not be included in any solution interval (even though this is an "or equal to" inequality), but I need the value in order to figure out what my intervals are. In this case, my intervals are (negative infinity, 3), (3, 6], and [6, positive infinity). Note the use of brackets to indicate that 6 can be included in the solution, but that 3 cannot. Using the Test-Point Method, I would pick a point in each interval and test for the sign on the result. I could use, say, x = 0, x = 4, and x = 7. Using the Factor Method, I solve each factor: –x + 6 > 0 for –x > –6, or x < 6; x – 3 > 0 for x > 3. Then I do the grid:
...fill in the signs on the factors:
...and solve for the sign on the rational function:
So the solution is all x's in the intervals (negative infinity, 3) and [6, positive infinity). There is another way to solve inequalities. You still have to find the zeroes (x-intercepts) first, but then you graph the function, and just look: wherever the graph is above the x-axis, the function is positive; wherever it is below the axis, the function is negative. For instance, for the first quadratic exercise, y = x2 – 3x + 2 > 0, we found the zeroes at x = 1 and x = 2. Now look at the graph:
On the graph, the solution is obvious: you would take the two intervals (but not the interval endpoints) where the line is above the x-axis. Or that huge polynomial we did at the top of this page: x5 + 3x4 – 23x3 – 51x2 + 94x + 120 > 0. We found the zeroes of the polynomial, being x = –5, x = –3, x = –1, x = 2, and x = 4. Now look at the graph:
On the picture, the solution is obvious: take the three intervals (together with the interval endpoints) where the line is above the x-axis. As you can probably guess, a graphing calculator can save you a lot of time on these inequalities — if you understand what you're doing. You should still show your work and reasoning, but don't be shy about using the pictures to confirm the algebra. << Previous Top | 1 | 2 | 3 | Return to Index
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