Chapter Closure
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Exercise 131 Page 251
a
Create a common denominator, then add the rational expressions. Can you find any common factors?
b
Create a common denominator, then subtract the rational expressions. Can you find any common factors?
c
Find common factors in the numerator and the denominator.
d
Create a common denominator, then add the rational expressions. Can you find any common factors?
a
See solution.
b
1/x-1, x≠ 1
c
4(x+3)/x-4, x≠ - 2, 0, or 4
d
1/4x, x≠ 2, 0
a
Before we add the fractions, let us think about forbidden values of x. Since we cannot divide by 0, the expressions in the denominators cannot be 0. Notice that the square of a number is always nonnegative. Therefore, x^2+9 must always be positive.
x^2+9>0
5x/x+3 + 3+x/x^2+9
ExpandFrac
a/b=a * ( x^2+9 )/b * ( x^2+9 )
5x( x^2+9 )/(x+3)( x^2+9 ) + 3+x/x^2+9
ExpandFrac
a/b=a * (x+3)/b * (x+3)
5x( x^2+9 )/(x+3)( x^2+9 ) + (3+x)(x+3)/( x^2+9 ) (x+3)
AddFrac
Add fractions
5x( x^2+9 )+(3+x)(x+3)/(x+3)( x^2+9 )
▼
Simplify numerator
Distr
Distribute 5x
5x^3+45x+(3+x)(x+3)/(x+3)( x^2+9 )
CommutativePropAdd
Commutative Property of Addition
5x^3+45x+(x+3)(x+3)/(x+3)( x^2+9 )
ProdToPowTwoFac
a* a=a^2
5x^3+45x+(x+3)^2/(x+3)( x^2+9 )
ExpandPosPerfectSquare
(a+b)^2=a^2+2ab+b^2
5x^3+45x+x^2+6x+9/(x+3)( x^2+9 )
AddTerms
Add terms
5x^3+51x+x^2+9/(x+3)( x^2+9 )
CommutativePropAdd
Commutative Property of Addition
5x^3+x^2+51x+9/(x+3)( x^2+9 )
b
Similarly as in Part A, we know that the denominator cannot be 0. Therefore we know that x≠ 1. To subtract the 1, we have to rewrite it as a fraction with a common denominator to the given fraction.
c
We want to multiply the given rational expressions.
x^2+5x+6/x^2-4x*4x/x+2
Before we start simplifying, we should figure out which values of x are excluded. In the second fraction, we immediately see that if x= -2, the expression in the denominator becomes 0.
4x/-2+2 ⇔ 4x/0
x^2-4x=0
▼
Solve for x
SplitIntoFactors
Split into factors
x* x-4* x=0
FactorOut
Factor out x
x(x-4)=0
ZeroProdProp
Use the Zero Product Property
lx=0 x-4=0
AddEqn
(II): LHS+4=RHS+4
lx_1=0 x_2=4
The expression is also undefined for x=0 and x=4. Let's start by multiplying the fractions.
x^2+5x+6/x^2-4x* 4x/x+2
MultFrac
Multiply fractions
(x^2+5x+6)4x/(x^2-4x)(x+2)
SplitIntoFactors
Split into factors
(x^2+5x+6)4x/(x* x-4* x)(x+2)
FactorOut
Factor out x
(x^2+5x+6)4x/x(x-4)(x+2)
To proceed, we must factor the expression in the numerator. To do that, we will have a look at what happens when we multiply (x+a) by (x+b). rc Expression & (x+ a)(x+ b) &⇓ Multiply & x^2+ ax+ bx+a b &⇓ Factor out x & x^2+( a+ b)x+ a b If we can rewrite x^2+5x+6 to match the last row, we can factor it.
x^2+5x+6
SplitIntoFactors
Split into factors
x^2+5x+(2)(3)
WriteSum
Write as a sum
x^2+( 2+ 3)x+( 2)( 3)
Write as (x+ a)(x+ b)
(x+ 2)(x+ 3)
Let's replace (x^2+5x+6) in the numerator with the factored expression and simplify.
(x+2)(x+3)4x/x(x-4)(x+2)
ReduceFrac
a/b=.a /(x+2)./.b /(x+2).
(x+3)4x/x(x-4)
ReduceFrac
a/b=.a /x./.b /x.
(x+3)4/x-4
Distr
Distribute 4
4x+12/x-4
d
We want to divide the given rational expressions.
x^2-2x/x^2-4x+4÷4x^2/x-2
Like in previous parts, we want to identify where the function is undefined. We can immediately see that the second fraction is undefined when x=2 since this gives 0 in its denominator. To determine when the first fraction is undefined, we will set it equal to 0.
x^2-4x+4=0
▼
Factor
WritePow
Write as a power
x^2-4x+2^2=0
SplitIntoFactors
Split into factors
x^2-2(x)(2)+2^2=0
FacNegPerfectSquare
a^2-2ab+b^2=(a-b)^2
(x-2)^2=0
x=2
Here we have figured out two things. The expression can be factored as (x-2)^2 and it is not defined when x=2. Let's replace x^2-4x+4 in the first fraction by the factored expression and then start simplifying the expression.
x^2-2x/(x-2)^2÷ 4x^2/x-2
DivFracByFracDiv
a/b÷c/d=a/b*d/c
x^2-2x/(x-2)^2* x-2/4x^2
MultFrac
Multiply fractions
(x^2-2x)(x-2)/(x-2)^2 4x^2
▼
Factor
x(x-2)(x-2)/(x-2)^2 4x^2
ProdToPowTwoFac
a* a=a^2
x(x-2)^2/(x-2)^2 4x^2
ReduceFrac
a/b=.a /x./.b /x.
(x-2)^2/(x-2)^2 4x
ReduceFrac
a/b=.a /(x-2)^2./.b /(x-2)^2.
1/4x
Having simplified the expression, we also see that x cannot equal 0.
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