- b.To begin, we must make certain that the
numerator and denominator are both com-
pletely factored and that all common terms are
canceled:
y 82 – –^6 y^4 = (y–(88)–(yy+8)) = (y––(^8 y)(–y 8 +)8)= –(y+ 8)
Now, determine the y-values that make this
expression equal to zero or undefined. Then,
we assess the sign of the expression on the left
side on each subinterval formed using these
values. To this end, observe that these values
are y= –8, 8. Next, we form a number line,
choose a real number in each subinterval, and
record the sign of the expression above each,
as follows:The inequality includes “equals,” we include
those values from the number line that make
the numerator equal to zero. We conclude that
the solution set is [–8, 8)∪(8,∞).- a.First, we must make certain that the numer-
ator and denominator are both completely fac-
tored and that all common terms are canceled:
= = =Next, determine the x-values that make this
expression equal to zero or undefined. Then,
we assess the sign of the expression on the left
side on each subinterval formed using these
values. To this end, observe that these values
are x= –8, 8. Now, we form a number line,
choose a real number in each subinterval, and
record the sign of the expression above each:Since the inequality does not include “equals,”
we do not include those values from the num-
ber line that make the numerator equal to
zero. Therefore, the solution set is (8,∞).- a.To begin, we must make certain that the
numerator and denominator are both com-
pletely factored and that all common terms
are canceled:
= =
=Next determine the x-values that make this
expression equal to zero or undefined. Then,
we assess the sign of the expression on the left
side on each subinterval formed using these
values. To this end, observe that these values
are x= –^12 ,–^25 , 1. Now, we form a number line,
choose a real number in each subinterval, and
record the sign of the expression above each:The inequality includes “equals,” we include
those values from the number line that make
the numerator equal to zero. We conclude that
the solution set is (–^12 ,–^25 )∪(–^25 , 1].-^12 –^251
+ – – +x– 1
2 x+ 1
(x– 1)(5x+ 2)(x– 1)
(x– 1)(5x+2)(2x+ 1)(x– 1)(5x^2 – 3x– 2)
(x– 1)(10x^2 + 9x+2)
^5 x^2 (x– 1) – 3x(x–1) – 2(x– 1)
10 x^2 (x– 1) + 9x(x– 1 +2(x– 1)8 8
– +
^1
x– 8
x(x+ 8)
x(x+ 8)(x– 8)
x(x+ 8)
x(x^2 – 64)
x^2 + 8x
x^2 – 64x8 8+ – –ANSWERS & EXPLANATIONS–