408 CHAPTER 3 Polynomial and Rational Functions 3-116
College Algebra—
This statement simply reinforces the idea that if a is
left of b on the number line, then As shown in
the diagram, and from which
for any x.
B. The second idea reiterates well-known ideas regarding
the multiplication of signed numbers. For any number
of factors:
if there are an even number of negative factors, the
result is positive;
if there are an odd number of negative factors, the
result is negative.
These two ideas work together to solve inequalities
using what we’ll call the push principle. Consider the
inequality The factored form is
and we want the product of these two
factors to be positive. From (A), both factors will be pos-
itive if is positive, since it’s the smaller of the two;
and both factors will be negative if since it’s
the larger. The solution set is found by solving these two
simple inequalities: gives and
gives If the inequality were
instead, we require one negative
factor and one positive factor. Due to order relations and
the number line, the larger factor must be the positive one:
so . The smaller factor must be the
negative one: and This gives the solu-
tion as can be verified using any alternative
method. Solutions to all other polynomial and rational
inequalities are an extension of these two cases.
Illustration 1
Solve using the push
principle.
Solution
The polynomial can be factored using the
tests for 1 and and synthetic division. The factors are
which we’ve conveniently
written in increasing order. For the product of three factors
to be negative we require: (1) three negative factors or (2) one
negative and two positive factors. The first condition is met
1x 221x 121x 326 0,
1
x
3
7x 6 6 0
3 6 x 6 4
x 6 4.x 4 6 0
x 7 3x 3 7 0
1x 421x 326 0
x 6 3.x 3 6 0
x 7 4x 4 7 0
x 3 6 0,
1x 42
1x 421x 327 0
x
2
x 12 7 0.
x 4 6 x 3
x 6 x 3,x 4 6 x
a 6 b.
by simply making the largest factor negative, as it will
ensure the smaller factors are also negative: so
The second condition is met by making the
smaller factor negative and the “middle” factor positive:
and The second solution interval
is and or
Note the push principle does not require the testing of
intervals between the zeroes, nor the “cross/bounce”
analysis at the zeroes and vertical asymptotes (of rational
functions). In addition, irreducible quadratic factors can
still be ignored as they contribute nothing to the solution
of real inequalities, and factors of even multiplicity can be
overlooked precisely because there is no sign change at
these roots.
Illustration 2
Solve
using the push principle.
Solution
Since the factor does not affect the
solution set, this inequality will have the same solution as
Further, since will be non-
negative for all x, the original inequality has the same solu-
tion set as (x 3) 0! The solution is
With some practice, the push principle can be a very
effective tool. Use it to solve the following exercises.
Check all solutions by graphing the function on a graph-
ing calculator.
Exercise 1:
Exercise 2:
Exercise 3:
Exercise 4:
Exercise 5:
Exercise 6: 1x
2
521x
2
921x 22
2
1x 12 0
x
4
x
2
12 7 0
x
3
3x 2 0
x
3
13x 12 6 0
x 1
x
2
4
7 0
x
3
3x 18 0
x 3.
1x 22
2
1x 22
2
1x 32 0.
1x
2
12
1x
2
121x 22
2
1x 32 0
1 6 x 6 2.x 7 1,x 6 2
x 1 7 0.x 2 6 0
x 6 3.
x 3 6 0
CUMULATIVE REVIEW CHAPTERS R–3
1. Solve for
2. Solve for
3. Factor the expressions:
a. b.
4. Solve using the quadratic formula. Write answers
in both exact and approximate form:
2x
2
4x 1 0.
x
3
3x
2
4x 12x
3
1
x:
2
x 1
1
5
x
2
1
R:
1
R
1
R
1
1
R
2
5. Solve the following inequality: or
6. Name the eight toolbox functions, give their
equations, then draw a sketch of each.
7. Use substitution to verify that is a
solution to
8. Solve the rational inequality:
x 4
x 2
6 3.
x
2
4x 13 0.
x 2 3i
5 x 6 4.
x 3 6 5
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