Algebra Rules Review Sheet Page 16
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16
REVIEW OF ALGEBRA
1
c − 1 + 1
c
1 +
c − 1
c
c
c − 1
c − 1
c − 1
27.
=
=
=
·
=
1
c − 1 − 1
c − 2
c − 2
c − 1
c − 2
1 −
c − 1
c − 1
c − 1
1
1
1 + x
2 + x + 1 + x
3 + 2x
28. 1 +
= 1 +
= 1 +
=
=
1
1 + x + 1
2 + x
2 + x
2 + x
1 +
1 + x
1 + x
29. 2x + 12x
3
2
2
= 2x · 1 + 2x · 6x
= 2x(1 + 6x
)
30. 5ab − 8abc = ab · 5 − ab · 8c = ab(5 − 8c)
31. The two integers that add to give 7 and multiply to give 6 are 6 and 1. Therefore x
2
+ 7x + 6 = (x + 6)(x + 1).
32. The two integers that add to give −1 and multiply to give −6 are −3 and 2.
Therefore x
2
− 2x − 6 = (x − 3)(x + 2).
33. The two integers that add to give −2 and multiply to give −8 are −4 and 2.
Therefore x
2
− 2x − 8 = (x − 4)(x + 2).
34. 2x
2
+ 7x − 4 = (2x − 1)(x + 4)
35. 9x
2
2
− 4) = 9(x − 2)(x + 2) [Equation 3 with a = x, b = 2]
− 36 = 9(x
36. 8x
2
+ 10x + 3 = (4x + 3)(2x + 1)
37. 6x
2
− 5x − 6 = (3x + 2)(2x − 3)
38. x
2
2
[Equation 1 with a − x, b = 5]
+ 10x + 25 = (x + 5)
39. t + 1 = (t + 1)(t
3
2
− t + 1) [Equation 5 with a = t, b = 1]
40. 4t
2
2
2
2
= (2t − 3s)(2t + 3s) [Equation 3 with a = 2t, b = 3s]
− 9s
= (2t)
− (3s)
41. 4t
2
2
[Equation 2 with a = 2t, b = 3]
− 12t + 9 = (2t − 3)
42. x
3
2
+ 3x + 9) [Equation 4 with a = x, b = 3]
− 27 = (x − 3)(x
43. x
3
2
2
2
[Equation 1 with a = x, b = 1]
+ 2x
+ x = x(x
+ 2x + 1) = x(x + 1)
44. Let p(x) = x
3
2
+ 5x − 2, and notice that p(1) = 0, so by the Factor Theorem, (x − 1) is a factor.
− 4x
Use long division (as in Example 8):
2
x
− 3x + 2
3
2
x − 1 x
− 4x
+ 5x − 2
3
2
x
− x
2
− 3x
+ 5x
2
− 3x
+ 3x
2x − 2
2x − 2
Therefore x
3
2
2
2
− 4x
+ 5x − 2 = (x − 1)(x
− 3x + 2) = (x − 1)(x − 2)(x − 1) = (x − 1)
(x − 2).
45. Let p(x) = x
3
2
− x − 3, and notice that p(1) = 0, so by the Factor Theorem, (x − 1) is a factor.
+ 3x
Use long division (as in Example 8):
2
x
+ 4x + 3
3
2
x − 1 x
+ 3x
− x − 3
3
2
x
− x
2
4x
− x
2
4x
− 4x
3x − 3
3x − 3
Therefore x
3
2
2
+ 3x
− x − 3 = (x − 1)(x
+ 4x + 3) = (x − 1)(x + 1)(x + 3).
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