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6e reviewofalgebra.pdf


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REVIEW OF ALGEBRA ■ 3

x
x⫹y
⫹1
y
y
x⫹y
x
x共x ⫹ y兲
x 2 ⫹ xy
(d)





y
x⫺y
y
x⫺y
y共x ⫺ y兲
xy ⫺ y 2
1⫺
x
x
FACTORING

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 ⫹ bx ⫹ c we note that
共x ⫹ r兲共x ⫹ s兲 苷 x 2 ⫹ 共r ⫹ s兲x ⫹ rs
so we need to choose numbers r and s so that r ⫹ s 苷 b and rs 苷 c.
EXAMPLE 4 Factor x 2 ⫹ 5x ⫺ 24.
SOLUTION The two integers that add to give 5 and multiply to give ⫺24 are ⫺3 and 8.
Therefore

x 2 ⫹ 5x ⫺ 24 苷 共x ⫺ 3兲共x ⫹ 8兲
EXAMPLE 5 Factor 2x 2 ⫺ 7x ⫺ 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 苷 ⫺4. Experimentation reveals that
2x 2 ⫺ 7x ⫺ 4 苷 共2x ⫹ 1兲共x ⫺ 4兲

Stewart: Calculus, Sixth Edition. ISBN: 0495011606. © 2008 Brooks/Cole. All rights reserved.

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

a 2 ⫺ b 2 苷 共a ⫺ b兲共a ⫹ b兲

The analogous formula for a difference of cubes is
4

a 3 ⫺ b 3 苷 共a ⫺ b兲共a 2 ⫹ ab ⫹ b 2 兲

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

a 3 ⫹ b 3 苷 共a ⫹ b兲共a 2 ⫺ ab ⫹ b 2 兲

EXAMPLE 6

(a) x 2 ⫺ 6x ⫹ 9 苷 共x ⫺ 3兲2
(b) 4x 2 ⫺ 25 苷 共2x ⫺ 5兲共2x ⫹ 5兲
(c) x 3 ⫹ 8 苷 共x ⫹ 2兲共x 2 ⫺ 2x ⫹ 4兲

(Equation 2; a 苷 x, b 苷 3 )
(Equation 3; a 苷 2x, b 苷 5 )
(Equation 5; a 苷 x, b 苷 2 )