Repaso Algebra
Páginas: 6 (1370 palabras)
Publicado: 13 de febrero de 2013
Review of Algebra
2
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REVIEW OF ALGEBRA
Review of Algebra
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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 band so b c b c ba c (Commutative Law) (Associative Law) (Distributive law)
ab c
a bc
1 in the Distributive Law, we get c 1 b c 1b 1c
EXAMPLE 1
(a) 3xy 4x 3 4 x 2y 12x 2y (b) 2t 7x 2tx 11 14tx 4t 2x 22t (c) 4 3 x 2 4 3x 6 10 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 onefactor by each term in the other factor and adding the products. Schematically, we have a In the case where c or
1
b c
d
a and d a b
b, we have
2
a2
ba
ab
b2
a
b
2
a2
2ab
b2
Similarly, we obtain
2
a
b
2
a2
2ab
b2
REVIEW OF ALGEBRA
x
3
EXAMPLE 2
6x 2 3x (a) 2x 1 3x 5 (b) x 6 2 x 2 12x 36 2x 6 (c) 3 x 1 4x 3
10x 3 4x2 12x 2 12x 2
5 x 3x 5x
6x 2
7x
5 12 12
3 2x 9 2x 21
Fractions
To add two fractions with the same denominator, we use the Distributive Law: a b Thus, it is true that a b c a b c b c b 1 b a 1 b c 1 a b c a b c
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 differentdenominators, we use a common denominator: a b c d ad bd bc
We multiply such fractions as follows: a b In particular, it is true that a b a b a b c d ac bd
To divide two fractions, we invert and multiply: a b c d
a b
d c
ad bc
4
s
REVIEW OF ALGEBRA
EXAMPLE 3
3 x x 3 x 3x 2 xx 1 3x 6 x 2 x (b) x 1 x 2 x 1 x 2 x2 x 2 2 x 2x 6 2 x x 2 s2t ut s 2 t 2u s2t 2 (c) u 2 2u 2 xx y 1 y y x xx y x 2 xy x y (d) x y y x y yx y xy y 2 y 1 x x (a) 1
Factoring
x
3
x x
3 x
We have used the Distributive Law to expand certain algebraic expressions. We sometimes 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 x r x s
bx x2
c we note that r s sx rs c.
so we need to choose numbers r and s so that r
EXAMPLE 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 4.
24
x
3 x
8
7x
SOLUTIONEven 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
4. Experimentation reveals that 4 2x 1 x 4
Some special 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
REVIEW OF ALGEBRA
x
5
The analogous formulafor 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
EXAMPLE 6
(a) x 2 6x 9 x 32 2 (b) 4x 25 2x 5 2x 5 (c) x 3 8 x 2 x 2 2x 4 x2 x
2
(Equation 2; a (Equation 3; a (Equation 5; a
x, b 3) 2x, b 5) x, b 2)
EXAMPLE 7 Simplify
16 . 2x 8
SOLUTIONFactoring 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 3 or more, we sometimes use the following fact.
6 The Factor Theorem If P is a polynomial and P b
0, then x
b is a factor
of P x .
EXAMPLE 8 Factor x 3
3
3x 2
2
10x
24.
SOLUTION Let P x x 3x 10x 24. If P b 0, where b is an integer,...
2
s
REVIEW OF ALGEBRA
Review of Algebra
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
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 band so b c b c ba c (Commutative Law) (Associative Law) (Distributive law)
ab c
a bc
1 in the Distributive Law, we get c 1 b c 1b 1c
EXAMPLE 1
(a) 3xy 4x 3 4 x 2y 12x 2y (b) 2t 7x 2tx 11 14tx 4t 2x 22t (c) 4 3 x 2 4 3x 6 10 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 onefactor by each term in the other factor and adding the products. Schematically, we have a In the case where c or
1
b c
d
a and d a b
b, we have
2
a2
ba
ab
b2
a
b
2
a2
2ab
b2
Similarly, we obtain
2
a
b
2
a2
2ab
b2
REVIEW OF ALGEBRA
x
3
EXAMPLE 2
6x 2 3x (a) 2x 1 3x 5 (b) x 6 2 x 2 12x 36 2x 6 (c) 3 x 1 4x 3
10x 3 4x2 12x 2 12x 2
5 x 3x 5x
6x 2
7x
5 12 12
3 2x 9 2x 21
Fractions
To add two fractions with the same denominator, we use the Distributive Law: a b Thus, it is true that a b c a b c b c b 1 b a 1 b c 1 a b c a b c
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 differentdenominators, we use a common denominator: a b c d ad bd bc
We multiply such fractions as follows: a b In particular, it is true that a b a b a b c d ac bd
To divide two fractions, we invert and multiply: a b c d
a b
d c
ad bc
4
s
REVIEW OF ALGEBRA
EXAMPLE 3
3 x x 3 x 3x 2 xx 1 3x 6 x 2 x (b) x 1 x 2 x 1 x 2 x2 x 2 2 x 2x 6 2 x x 2 s2t ut s 2 t 2u s2t 2 (c) u 2 2u 2 xx y 1 y y x xx y x 2 xy x y (d) x y y x y yx y xy y 2 y 1 x x (a) 1
Factoring
x
3
x x
3 x
We have used the Distributive Law to expand certain algebraic expressions. We sometimes 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 x r x s
bx x2
c we note that r s sx rs c.
so we need to choose numbers r and s so that r
EXAMPLE 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 4.
24
x
3 x
8
7x
SOLUTIONEven 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
4. Experimentation reveals that 4 2x 1 x 4
Some special 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
REVIEW OF ALGEBRA
x
5
The analogous formulafor 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
EXAMPLE 6
(a) x 2 6x 9 x 32 2 (b) 4x 25 2x 5 2x 5 (c) x 3 8 x 2 x 2 2x 4 x2 x
2
(Equation 2; a (Equation 3; a (Equation 5; a
x, b 3) 2x, b 5) x, b 2)
EXAMPLE 7 Simplify
16 . 2x 8
SOLUTIONFactoring 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 3 or more, we sometimes use the following fact.
6 The Factor Theorem If P is a polynomial and P b
0, then x
b is a factor
of P x .
EXAMPLE 8 Factor x 3
3
3x 2
2
10x
24.
SOLUTION Let P x x 3x 10x 24. If P b 0, where b is an integer,...
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