Since the inequality does not include “equals,”
we do not include those values from the num-
ber line that make the polynomial equal to zero.
Therefore, the solution set is (–∞, –6)∪(6,∞).
- b.Determine the x-values that make the
expression on the left side equal to zero. First,
factor the polynomials:
9 x^2 – 25 = (3x)^2 – (5)^2 = (3x– 5)(3x+ 5)
Next, set each factor equal to zero and solve
for xto conclude that the zeros of the polyno-
mial are –^53 and ^53 . Now, assess the sign of the
expression on the left side on each subinterval
formed using these values. To this end, 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
those values from the number line that make
the polynomial equal to zero. The solution set
is [–^53 ,^53 ].
- c.Determine the x-values that make the
expression on the left side equal to zero. Begin
by factoring the polynomial, if possible. How-
ever, note that 5x^2 + 49 cannot be factored fur-
ther. Moreover, since both terms are positive
for any value ofx, the sum is positive for every
value ofx. Therefore, the solution set is the
empty set.
452. a.Find the x-values that make the expression
on the left side equal to zero. First, factor the
polynomial:
6 x^2 – 24 = 6(x^2 ) – 6(4) = 6(x^2 –4) =
6((x)^2 – (2)^2 ) = 6(x–2)(x+ 2)
Next, set each factor equal to zero. Solve for x
to find that the zeros of the polynomial: –2
and 2. Now, 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:
Since the inequality includes “equals,” we
include those values from the number line that
make the polynomial equal to zero. The solu-
tion set is (–∞, –2]∪[2,∞).
- a.The strategy is to determine the x-values
that make the expression on the left side equal
to zero. Doing so requires that we first factor
the polynomial:
5 x(2x+ 3) – 7(2x+ 3) = (2x+ 3)(5x+ 7)
Set each factor equal to zero and solve for x.
The zeros of the polynomial: –^32 and ^75 .Next
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 duly formed subintervals, and
record the sign of the expression above each:
3
2
7
5
+ – +
22
+ – +
5
3
5
3
+ – +
ANSWERS & EXPLANATIONS–
