Lesson Solving inequalities for high degree polynomials factored into a product of linear binomials

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Solving inequalities for high degree polynomials factored into a product of linear binomials

In this lesson you will learn how to solve inequalities for high degree polynomials that are factored into a product of linear binomials. Typical examples are

1) %28x-1%29%2A%28x-2%29%2A%28x-3%29

and 2)

%28x-1%29%2A%28x-2%29%2A%28x-3%29%2A%28x-4%29

Problem 1

Solve the inequality %28x-1%29%2A%28x-2%29%2A%28x-3%29

.

Solution

The left side is the product of three linear factors %28x-1%29

and

.
If

, then all three factors are negative, so their product is negaive:
%28x-1%29%2A%28x-2%29%2A%28x-3%29

.

If

, then the factor x-1

is non-negative, while the factors x-2

and

are non-positive, so the product of three is non-negative: %28x-1%29%2A%28x-2%29%2A%28x-3%29

.

If

, then two factors x-1

and

are positive, while the factor x-3

is negative, so the product of three is negative: %28x-1%29%2A%28x-2%29%2A%28x-3%29

.

If

, then all three factors are non-negative, so their product is non-negaive:
%28x-1%29%2A%28x-2%29%2A%28x-3%29

Fig.1. The factor x-1

and

plots

Fig.2. The polynomial

plot

Thus the polynomial %28x-1%29%2A%28x-2%29%2A%28x-3%29

is non-negative at

and

. It is the set of all real numbers satisfying the given inequality %28x-1%29%2A%28x-2%29%2A%28x-3%29

.

Figure 1 shows the plots of the linear factors x-1

and

separately. Figure 2 shows the plot of the product of these factors %28x-1%29%2A%28x-2%29%2A%28x-3%29

.
It is clearly seen from the Figure 2 that the given polynomial %28x-1%29%2A%28x-2%29%2A%28x-3%29

is non-negative on the segment

and on the semi-infinite interval

.

Answer. The solution of the given inequality %28x-1%29%2A%28x-2%29%2A%28x-3%29

is the union of the segment

and the semi-infinite interval

Problem 2

Solve yourself the inequality %28x%2B1%29%2A%28x-2%29%2A%28x-3%29

Problem 3

Solve yourself the inequality -%28x%2B1%29%2A%28x%2B2%29%2A%28x-3%29

Problem 4

Solve the inequality %28x-1%29%2A%28x-2%29%2A%28x-3%29%2A%28x-4%29

.

Solution

The left side is the product of four linear factors %28x-1%29

and

.
If

, then all four factors are non-positive, so their product is non-negative:
%28x-1%29%2A%28x-2%29%2A%28x-3%29%2A%28x-4%29

.

If

, then the factor x-1

is positive, while the factors x-2

and

are negative. So the product of four is negative: %28x-1%29%2A%28x-2%29%2A%28x-3%29%2A%28x-4%29

.

If

, then two factors x-1

and

are non-positive, while the two
other factors x-3

and

are non-negative. So the product of four is non-negative:
%28x-1%29%2A%28x-2%29%2A%28x-3%29%2A%28x-4%29

.

If

, then three factors x-1

and

are positive, while the
factor x-4

is negative. So the product of four is negative:
%28x-1%29%2A%28x-2%29%2A%28x-3%29%2A%28x-4%29

Fig.3. The factor x-1

and

plots

Fig.4. The polynomial

plot

, then all four factors are non-negative, so their product is non-negaive: %28x-1%29%2A%28x-2%29%2A%28x-3%29%2A%28x-4%29

.

Thus the polynomial %28x-1%29%2A%28x-2%29%2A%28x-3%29%2A%28x-4%29

is non-negative at

and

. It is the set of all real numbers satisfying the given inequality.

Figure 3 shows the plots of the linear factors x-1

and

separately. Figure 4 shows the plot of the product of these factors %28x-1%29%2A%28x-2%29%2A%28x-3%29%2A%28x-4%29

.
It is clearly seen from the Figure 4 that the given polynomial %28x-1%29%2A%28x-2%29%2A%28x-3%29%2A%28x-4%29

is non-negative on the segment

and on the segment

.

Answer. The solution of the given inequality %28x-1%29%2A%28x-2%29%2A%28x-3%29%2A%28x-4%29

is the union of the segment

and the segment

Summary

To solve an inequality with a polynomial on the left side factored to a product of linear binomials, analyze the sign of each factor and the sign of their product
by moving along the number line from left to right.

My other lessons on solving inequalities are
    - Solving simple and simplest linear inequalities
    - Solving absolute value inequalities
    - Advanced problems on solving absolute value inequalities
    - Solving systems of linear inequalities in one unknown
    - Solving compound inequalities

    - What number is greater? Comparing magnitude of irrational numbers
    - Arithmetic mean and geometric mean inequality
    - Arithmetic mean and geometric mean inequality - Geometric interpretations
    - Harmonic mean
    - Prove that if a, b, and c are the sides of a triangle, then so are sqrt(a), sqrt(b), and sqrt(c)

    - Solving problems on quadratic inequalities
    - Solving inequalities for rational functions with numerator and denominator factored into a product of linear binomials
    - Solving inequalities for rational functions with non-zero right side
    - Another way solving inequalities for rational functions with non-zero right side

    - Advanced problems on inequalities
    - Challenging problems on inequalities
    - Solving systems of inequalities in two unknowns graphically in a coordinate plane
    - Solving word problems on inequalities
    - Proving inequalities
    - Math circle level problem on inequalities
    - Math Olympiad level problems on inequalities
    - Entertainment problems on inequalities
under the topic Inequalities, trichotomy of the section Algebra-I.

My lessons on domains of functions are
    - Domain of a function which is a quadratic polynomial under the square root operator
    - Domain of a function which is a high degree polynomial under the square root operator
    - Domain of a function which is the square root of a rational function.
under the topic Functions, Domain of the section Algebra-I.

See also OVERVIEW of lessons on inequalities and domains of functions.

Use this file/link ALGEBRA-I - YOUR ONLINE TEXTBOOK to navigate over all topics and lessons of the online textbook ALGEBRA-I.

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