Since the inequality does not include “equals,”
do not include those values from the number
line that make the polynomial equal to zero. As
such, the solution set is(–∞,–^32 )∪(^75 ,∞).
- c.Find the x-values that make the expression
on the left side equal to zero. First, factor the
polynomial:
5 x(^23 x+ 7) – (^23 + 7) = (5x– 1)(^23 x+ 7)
Next, set each factor equal to zero and solve
for x.The zeros of the polynomial are ^15 and
- ^221 . Assess the sign of the expression on the
left side on each subinterval formed using
these values. Form a number line, choose a
real number in each of the subintervals, and
record the sign of the expression above each:
The inequality includes “equals,” so we include
those values from the number line that make
the polynomial equal to zero. The solution set
is [–^221 ,^15 ].
- c.Determine the x-values that make the
expression on the left side equal to zero. To do
this, factor the polynomial:
28 x(5 – x) – 7x^3 (5 – x) = (28x– 7x^3 )(5 – x) =
7 x(4 – x^2 )(5 – x)
= 7x(2^2 – x^2 )(5 – x) = 7x(2 – x)(2 + x)(5 – x)
Next, set each factor equal to zero. Solve for x
to find the zeros of the polynomial, which are
0, –2, 2, and 5. Now, assess the sign of the
expression on the left side on each subinterval
formed using these values. To this end, we
form a number line, choose a real number in
each of the subintervals, and record the sign of
the expression above each:
Since the inequality includes “equals,” include
the values from the number line that make the
polynomial equal to zero. The solution set
is(–∞, –2)∪[0, 2]∪(5,∞).
456. c.First, determine the x-values that make the
expression on the left side equal to zero. Doing
so requires that we factor the polynomial:
75 x^4 + 30x^3 + 3x^2 = 3x^2 [25x^2 + 10x+ 1] =
3 x^2 [25x^2 + 5x+ 5x+ 1]
= 3x^2 [5x(5x+ 1) + (5x+ 1)] = 3x^2 [(5x+ 1) +
(5x+ 1)] = 3x^2 (5x+ 1)^2
Set each factor equal to zero, then solve for x
to find the zeros of the polynomial: 0 and –^15 .
Assess the sign of the expression on the left
side on each subinterval formed using these
values: Form a number line, choose a real
number in each subinterval, and record the
sign of the expression above each:
The inequality includes “equals,” so we include
those values from the number line that make
the polynomial equal to zero. Since every
x-value that is not a zero of the polynomial
results in a positive quantity, the solution set
consists of only the zeros of the polynomial,
namely {–^15 , 0}.
0
+ + +
1
2 0 2 5
++–
21
2
1
5
+ – +
ANSWERS & EXPLANATIONS–
