Math 135
Factoring and Rational Expressions
Solutions
1. Factor into linear factors whenever possible:
(a) x2 + 5x + 6 = 0
Answer 1.
x2 + 5x + 6 = 0
(x + 3)(x + 2) = 0
(b) x3 − 3x2 + 2x = 0
Answer 2.
x3 − 3x2 + 2x = 0
x(x2 − 3x + 2) = 0
x(x − 2)(x − 1) = 0
(c) x(x − 3) − 2x + 6 = 0
Answer 3.
x(x − 3) − 2x + 6 = 0
x(x − 3) − 2(x − 3) = 0
(x − 2)(x − 3) = 0
(d) x2 + 2x + 1 = 0
Answer 4.
x2 + 2x + 1 = 0
(x + 1)(x + 1) = 0
(x + 1)2 = 0
(e) 8x3 − 27 = 0
Answer 5.
8x3 − 27
(2x)3 − 33
(2x − 3)((2x)2 + 2x · 3 + 32 )
(2x − 3)(4x2 + 6x + 9)
=
=
=
=
0
0
0
0
The expression (4x2 + 6x + 9) does not factor at all. Later we will develop the
tools that will allow us to see this easily.
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R Spring - 2014
Math 135
Factoring and Rational Expressions
Solutions
2. Solve and write the solutions in set notation:
(a)
1
x+1
=3
Answer 6.
1
= 3
x+1
1
−3
x+1
1
x+1
−3·
x+1
x+1
1 − 3(x + 1)
x+1
−3x − 2
x+1
(b)
Thus,
1
x+1
−1
x−1
2x+1
2
+
= 0
= 0
= 0
= 0
= 3 provided that −3x − 2 = 0 and x 6= −1. The solution set is {− 23 }.
=0
Answer 7.
2x + 1
−1
+
x−1
2
−1 2 2x + 1 x − 1
· +
·
x−1 2
2
x−1
−2 + (2x + 1)(x − 1)
2(x − 1)
−2 + 2x2 − x − 1
2(x − 1)
2x2 − x − 3
2(x − 1)
(2x − 3)(x + 1)
2(x − 1)
−1
Thus, x−1
+ 2x+1
= 0 provided that x =
2
3
is {−1, 2 }.
University of Hawai‘i at Mānoa
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3
2
= 0
= 0
= 0
= 0
= 0
= 0
or x = −1 and x 6= 1. The solution set
R Spring - 2014
Math 135
Factoring and Rational Expressions
(c) 1 +
1
x+1
+
2
x−1
Solutions
=0
Answer 8.
1
2
+
x+1 x−1
(x + 1)(x − 1)
1
x−1
2
x+1
1·
+
·
+
·
(x + 1)(x − 1) x + 1 x − 1 x − 1 x + 1
(x + 1)(x − 1) + (x − 1) + 2(x + 1)
(x + 1)(x − 1)
x2 − 1 + x − 1 + 2x + 2
(x + 1)(x − 1)
x2 + 3x
(x + 1)(x − 1)
x(x + 3)
(x + 1)(x − 1)
1+
= 0
= 0
= 0
= 0
= 0
= 0
If x = ±1, then we are dividing by zero so these numbers, if they are solutions,
are invalid. The numerator and hence the entire expression is 0, whenever
x = 0 or x = −3. The solution set is {−3, 0}.
University of Hawai‘i at Mānoa
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R Spring - 2014