Mathematics
Grade 12
15 min
Solve higher degree inequalities
Solve higher degree inequalities
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1
Introduction & Learning Objectives
Learning Objectives
Identify the critical points of a higher degree polynomial by finding its real zeros.
Construct a sign analysis chart to determine the sign of a polynomial in intervals defined by its critical points.
Solve higher degree polynomial inequalities of the form P(x) > 0, P(x) < 0, P(x) ≥ 0, and P(x) ≤ 0.
Correctly interpret the role of multiplicity in determining whether the sign of the polynomial changes at a critical point.
Express the solution set of a higher degree inequality using proper interval notation.
Connect the algebraic solution of a polynomial inequality to the graphical representation of the function (i.e., where the graph is above or below the x-axis).
How can a company modeling its profit with a cubic function P(x) determine the producti...
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Key Concepts & Vocabulary
TermDefinitionExample
Higher Degree InequalityAn inequality that can be written in the form P(x) > 0, P(x) < 0, P(x) ≥ 0, or P(x) ≤ 0, where P(x) is a polynomial of degree 3 or greater.x³ - 4x² + x + 6 ≤ 0 is a higher degree inequality.
Critical PointsThe real roots or zeros of the polynomial P(x). These are the x-values where P(x) = 0, and they are the only points where the polynomial can change its sign.For the polynomial P(x) = (x-1)(x+2)(x-5), the critical points are x = 1, x = -2, and x = 5.
Test IntervalsThe intervals on the number line that are created by plotting the critical points. The sign of the polynomial is constant within each of these intervals.The critical points -2, 1, and 5 create the test intervals: (-∞, -2), (-2, 1), (1, 5), and (5, ∞).
Sign Analysis ChartA tabl...
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Core Formulas
The Test Point Method
1. Set P(x) = 0 to find critical points. \\ 2. Plot points on a number line to create test intervals. \\ 3. Choose a test value 'c' in each interval. \\ 4. Evaluate the sign of P(c). This sign applies to the entire interval.
This is the fundamental procedure for solving any polynomial inequality. It relies on the fact that a polynomial can only change sign at its real zeros.
Multiplicity Sign Change Rule
Let 'r' be a real root of P(x) with multiplicity 'm'. \\ If 'm' is odd, the sign of P(x) changes at x=r. \\ If 'm' is even, the sign of P(x) does not change at x=r.
This is a powerful shortcut for sign analysis. Once you determine the sign in one interval, you can use this rule to quickly determine the s...
4 more steps in this tutorial
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Challenging
Which of the following polynomial inequalities has the solution set (-∞, -1) U (4, ∞)?
A.(x + 1)(x - 4) < 0
B.(x + 1)(x - 4) > 0
C.(x - 1)(x + 4) > 0
D.(x - 1)(x + 4) < 0
Challenging
Solve the inequality x³ - 4x² + x + 6 ≤ 0. (Note: You may need the Rational Root Theorem to find the first root).
A.(-∞, -1] U [2, 3]
B.[-1, 2] U [3, ∞)
C.(-∞, 2] U [3, ∞)
D.[-1, 3]
Challenging
Consider the polynomial P(x) = (x-5)²(x+2). Let S1 be the solution set for P(x) > 0 and S2 be the solution set for P(x) ≥ 0. How do S1 and S2 differ?
A.S2 includes the point x = -2, but S1 does not.
B.S1 and S2 are identical.
C.S2 includes the points x = -2 and x = 5, but S1 includes neither.
D.S2 is the interval [-2, ∞), while S1 is the interval (-2, ∞).
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