3.8 simplifying rational expressions p. 161-164. vocabulary rational expression: ratio of 2...
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Excluded Value Set denominator = 0 (you know it can’t = 0) Set denominator = 0 (you know it can’t = 0) Solve for the variable Solve for the variable That is your excluded value. That is your excluded value.TRANSCRIPT
3.8 Simplifying Rational 3.8 Simplifying Rational ExpressionsExpressions
p. 161-164p. 161-164
VocabularyVocabulary Rational Expression: ratio of 2 polynomialsRational Expression: ratio of 2 polynomials
Excluded Value: once factored (before Excluded Value: once factored (before simplified), the value that would make the simplified), the value that would make the denominator 0denominator 0
Simplest Form: when numerator & Simplest Form: when numerator & denominator have no common factorsdenominator have no common factors
Excluded ValueExcluded Value Set denominator = 0 (you know it can’t = 0)Set denominator = 0 (you know it can’t = 0)
Solve for the variableSolve for the variable
That is your excluded value.That is your excluded value.
EXAMPLE 1 Find excluded values
Find the excluded values, if any, of the expression.
a. x + 810x
SOLUTION
a. The expression x + 8 is undefined when 10x = 0, or x = 0. 10x
ANSWER
The excluded value is 0.
EXAMPLE 1 Find excluded values
Find the excluded values, if any, of the expression.
b. 2y + 14
5
SOLUTION
The expression 5 is undefined when
2y + 14 = 0, or y = – 7.2y + 14
ANSWER
The excluded value is – 7.
EXAMPLE 1 Find excluded values
Find the excluded values, if any, of the expression.
c. v 2 – 9
4v
SOLUTION
c. The expression 4v is undefined when v2 – 9 = 0,
or (v + 3)(v – 3) = 0. The solutions of the equation are
– 3 and 3.
v2 – 9
The excluded values are – 3 and 3.
ANSWER
GUIDED PRACTICE for Example 1
Find the excluded values, if any, of the expression.
x + 23x – 5
1.
ANSWER
The excluded value is .53
3.
ANSWER
The excluded value is and 4 .32
–
n – 62n2 – 5n – 12
2. 2mm2 – 4
ANSWER
The excluded value is 2, and 2 .–
To simplifyTo simplify Factor numerator and denominatorFactor numerator and denominator
Cancel any common factors (anything over Cancel any common factors (anything over itself = 1)itself = 1)
EXAMPLE 2 Simplify expressions by dividing out monomials
Simplify the rational expression, if possible. State the excluded values.
a. r2r
SOLUTION
Divide out common factor.a. r2r = r
2r
= 12 Simplify.
ANSWER
The excluded value is 0.
EXAMPLE 2
Simplify the rational expression, if possible. State the excluded values.
b. 5x5(x + 2)
SOLUTION
b. 5x5(x + 2) = 5 x
5 (x + 2) Divide out common factor.
Simplify.= x(x + 2)
ANSWER
The excluded value is – 2.
Simplify expressions by dividing out monomials
EXAMPLE 2
Simplify the rational expression, if possible. State the excluded values.
SOLUTION
c. 6m3 – 12m2
18m2
c.18m2
6m3 – 12m2 = 6m2 (m – 2)6 3 m2
Factor numerator and denominator.
=6m2 (m – 2)6 3 m2 Divide out common factors.
= m – 2 3
Simplify.
ANSWER
The excluded value is 0.
Simplify expressions by dividing out monomials
EXAMPLE 2
Simplify the rational expression, if possible. State the excluded values.
SOLUTION
d. y7 – y
d. The expression y7 – y
is already in simplest form.
ANSWER
The excluded value is 7.
Simplify expressions by dividing out monomials
GUIDED PRACTICE for Example 2
5. 4 a3
22a62
11a3ANSWER The excluded value is 0.
6. 2c c + 5
2c c + 5ANSWER The excluded value is – 5.
7. 2s2 + 8s3s +12
ANSWER 2s3
The excluded value is – 4.
8. 8x8x3 + 16x2
ANSWER1
x2 + 2xThe excluded values are 0 and – 2.
EXAMPLE 3 Simplify an expression by dividing out binomials
Simplify x2 – 3x – 10x2 + 6x + 8
.State the excluded values.
SOLUTION
x2 – 3x – 10x2 + 6x + 8 = (x – 5)(x + 2)
(x + 4)(x + 2)Factor numerator and denominator.
(x – 5)(x + 2)(x + 4)(x + 2)
=
= x – 5x + 4
Divide out common factor.
Simplify.
ANSWER
The excluded values are – 4 and – 2.
EXAMPLE 3 Simplify an expression by dividing out binomials
CHECKCheck your simplification using a graphing calculator.
Graph y1x2 – 3x – 10x2 + 6x + 8 and y2= = x – 5
x + 4
The graphs coincide. So, the expressions are equivalent for all values of x other than the excluded values (– 4 and – 2).
EXAMPLE 4 Recognize opposites
Simplify x2 – 7x + 1216 – x2
.State the excluded values.
SOLUTION
x2 – 7x + 12 16 – x2 Factor numerator and
denominator.(x – 3)(x – 4)
– (x – 4)(4 + x)= Rewrite 4 – x as – ( x – 4).
Simplify.
– (x – 4)(4+ x)(x – 3)(x – 4)= Divide out common factor.
–(4 + x)(x – 3)=
(x – 3)(x – 4) (x – 4)(4 + x)=
ANSWERThe excluded values are – 4 and 4.
(x + 4)(x – 3)= –
GUIDED PRACTICE for Examples 3 and 4
Simplify the rational expression. State the excluded values.
9. x2 + 3x + 2x2 + 7x + 10
(x + 1)(x + 5)
ANSWER The excluded values are – 2 and – 5.
10.y2 – 64
y2 – 16y + 64 ANSWER
(y + 8)(y – 8)
The excluded value is 8
11. 5 + 4z – z2
z2 – 3z – 10ANSWER
(z + 1)(z + 2)
– The excluded values are 5 and – 2.
EXAMPLE 5 Simplify a rational model
46 – 2.2xC =
100 – 18x + 2.2x2
where x is the number of years since 1991. Rewrite the model so that it has only whole number coefficients.Then simplify the model.
The average cost C (in dollars per minute) for cell phone service in the United States during the period 1991–2000 can be modeled by
CELL PHONE COSTS
EXAMPLE 5 Simplify a rational model
C =46 – 2.2x
100 – 18x + 2.2x2Write model.
= 460 – 22x1000 – 180x + 22x2
Multiply numerator and denominator by 10.
Factor numerator and denominator.
=2(230 – 11x)
2(500 – 90x + 11x2)
Divide out common factor.
=2(230 – 11x)
2(500 – 90x + 11x2)
SOLUTION
= 230 – 11x
500 – 90x + 11x2Simplify.
GUIDED PRACTICE for Example 5
In Example 5, approximate the average cost per minute in 2000.
12.
ANSWERThe average cost per minute in 2000 is $.23/min.