Review of algebra

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REVIEW OF ALGEBRA
Here we review the basic rules and procedures of algebra that you need to know in order to be successful in calculus.
ARITHMETIC OPERATIONS

The real numbers have the following properties:

a b b a ab a b c a b ab c ab ac
In particular, putting a

ba c

ab c

a bc

(Commutative Law) (Associative Law) (Distributive law)

1 in the Distributive Law, we get b c 1 bc 1b 1c

and so

b
EXAMPLE 1

c

b

c

(a) 3xy (b) 2t 7x (c) 4

4x 2tx 2 3x

3 11 4

4 x 2y 14tx 3x 6

12x 2y 4t 2x 10 22t 3x

If we use the Distributive Law three times, we get

a

b c

d

a

bc

a

bd

ac

bc

ad

bd

This says that we multiply two factors by multiplying each term in one factor by each term in the other factor and adding theproducts. Schematically, we have

a
In the case where c or
1

b c

d

a and d a

b, we have b
2

a2

ba

ab

b2

a

b

2

a2

2ab

b2

Similarly, we obtain
2

a

b

2

a2

2ab

b2

EXAMPLE 2

Thomson Brooks-Cole copyright 2007

(a) 2x 1 3x 5 6x 2 3x 2 2 (b) x 6 x 12x 36 (c) 3 x 1 4x 3 2x 6

10x 3 4x 2 12 x 2

5 x 3x

6x 2 3 9

7x 2x 2x

5 1212

12x 2

5x

21

1

2 ■ REVIEW OF ALGEBRA

FRACTIONS

To add two fractions with the same denominator, we use the Distributive Law:

a b
Thus, it is true that

c b

1 b

a

1 b

c

1 a b

c

a
b

c

a

c

b

a b

c b

But remember to avoid the following common error:

|
b

a

c

a b

a c

(For instance, take a b c 1 to see the error.)To add two fractions with different denominators, we use a common denominator:

a b
We multiply such fractions as follows:

c d

ad

bc

bd

a b
In particular, it is true that

c d

ac bd

a b

a b

a b

To divide two fractions, we invert and multiply:
a b c d
EXAMPLE 3

a b

d c

ad bc

(a) (b)
Thomson Brooks-Cole copyright 2007

x
x

3

x x
x

3 x1

3 x

3
x 1

3x 2
x

2
s2t 2 2

x
ut 2

xx 1 1 x 2

3x
x
2

6
x

x2
2

x

x2 x2

2x x

6 2

(c)

s2t u

s 2 t 2u 2u

REVIEW OF ALGEBRA ■ 3

(d)

x y

1

x

y

y
x
x

x

y
y
x

x
y

1

y x

y

xx yx

y y

x2 xy

xy y2

FACTORING

We have used the Distributive Law to expand certain algebraic expressions. Wesometimes need to reverse this process (again using the Distributive Law) by factoring an expression as a product of simpler ones. The easiest situation occurs when the expression has a common factor as follows:
Expanding

3x(x-2)=3x@-6x
Factoring

To factor a quadratic of the form x 2

bx s

c we note that x2 r s sx rs c.

x

r x

so we need to choose numbers r and s so that rEXAMPLE 4 Factor x 2

b and rs

5x

24. 24 are 3 and 8.

SOLUTION The two integers that add to give 5 and multiply to give Therefore

x2
EXAMPLE 5 Factor 2x 2

5x

24

x

3 x

8

7x

4. 4. Experimentation reveals that 4 2x 1 x 4

SOLUTION Even though the coefficient of x 2 is not 1, we can still look for factors of the

form 2x

r and x

s, where rs 2x 2 7x

Somespecial quadratics can be factored by using Equations 1 or 2 (from right to left) or by using the formula for a difference of squares:
3

a2

b2

a

b a

b

The analogous formula for a difference of cubes is
4

a3

b3

a

b a2

ab

b2

which you can verify by expanding the right side. For a sum of cubes we have
5

a3

b3

a

b a2

ab

b2

Thomson Brooks-Colecopyright 2007

EXAMPLE 6

(a) x 2 6x 9 x 32 (b) 4x 2 25 2x 5 2x 5 (c) x 3 8 x 2 x 2 2x 4

(Equation 2; a (Equation 3; a (Equation 5; a

x, b 2x, b x, b

3) 5) 2)

4 ■ REVIEW OF ALGEBRA

EXAMPLE 7 Simplify

x2
x
2

16 . 2x 8

SOLUTION Factoring numerator and denominator, we have

x2
x
2

16 2x 8

x x

4 x 4 x

4 2

x x

4 2

To factor polynomials of degree...
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