Writing in Math, math homework help

1. OPEN ENDED Write two rational expressions that are equivalent.
2. OPEN ENDED Write two polynomials that have a LCM of d3 – d.

Writing in Math Use the information about rational expressions on page 462 to explain how rational expressions are used in mixtures. Include an example of a mixture problem that could be represented by

Find page 462 attached. 8-1
Multiplying and Dividing
Rational Expressions
GET READY for the Lesson
Main Ideas
Peanuts
• Simplify rational
expressions.
• Simplify complex
fractions.
Add
x lb of
peanuts.
New Vocabulary
The Goodie Shoppe sells candy
and nuts by the pound. One
item is a mixture made with
8 pounds of peanuts and
5 pounds of cashews.
8
8
13
mixture is peanuts. If the store
manager adds an additional
x pounds of peanuts to the
mixture, then
8 + x
13 + x
mixture will be peanuts.
Therefore, 3 5 or is of the
8 lb of peanuts
rational expression
complex fraction
5 lb of cashews
13 lb of nuts
NUTS
of the
8 + x
Simplify Rational Expressions A ratio of two polynomial expressions
such as
13 + x
is called a rational expression. Because variables in
algebra often represent real numbers, operations with rational numbers
and rational expressions are similar.
To write a fraction in simplest form, you divide both the numerator and
denominator by their greatest common factor (GCF). To simplify a
rational expression, you use similar techniques.
EXAMPLE
Simplify a Rational Expression
2x(x – 5)
a. Simplify
(x – 5)(x2 – 1)
Look for common factors.
1
2x(x – 5)
2x
(x – 5)(x2 – 1)
x2 – 1
1
2x
x2
1
x – 5
How is this similar to simplifying
10
.?
15
5
Simplify.
Study Tip
Excluded Values
Numbers that would
cause the expression
to be undefined are
called excluded
values.
b. Under what conditions is this expression undefined?
Just as with a fraction, a rational expression is undefined if
the denominator is equal to 0. To find when this expression
is undefined, completely factor the original denominator.
2x(x – 5)
2x(x – 5)
x2 – 1 = (x – 1)(x + 1)
(x – 5)(x2 – 1) (x – 5)(x – 1)(x + 1)
The values that would make the denominator equal to 0 are 5, 1, or
-1. So the expression is undefined when x = 5, x = 1, or x = -1.