Chapter Closure
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Exercise 113 Page 206
a
Factor the polynomials before multiplying the fractions.
b
Dividing fractions is the same as multiplying the first fraction with the reciprocal of the second.
c
Rewrite the number as a fraction with the same denominator as 3x+1.
d
Rewrite the fractions with the same denominator.
a
x-4/x+5
b
x^2+4/(x+3)(x+2)
c
6x+9/x+1
d
5/x+2
a
We want to multiply three given rational expressions.
Before multiplying the fractions, we want to factor the numerators and denominators. We will start with the numerator in the first fraction.
4x^2-13x+3
(x-3)(4x-1)
Let's continue with the denominator in the same fraction.
5x^2+23x-10
(x+5)(5x-2)
Let's continue by factoring the polynomials in the two remaining fractions.
x^2 + 6x - 27
(x+9)(x-3)
Only one more expression to factor!
x^2+5x-36
(x+9)(x-4)
We can now substitute the expressions with their factored form, multiply, and then simplify the fraction.
(x-3)(4x-1)/(x+5)(5x-2)*5x-2/(x+9)(x-3)*(x+9)(x-4)/4x-1
MultFrac
Multiply fractions
(x-3)(4x-1)(5x-2)(x+9)(x-4)/(x+5)(5x-2)(x+9)(x-3)(4x-1)
CommutativePropMult
Commutative Property of Multiplication
(x-3)(4x-1)(5x-2)(x+9)(x-4)/(x-3)(4x-1)(5x-2)(x+9)(x+5)
WriteProdFrac
Write as a product of fractions
(x-3)(4x-1)(5x-2)(x+9)/(x-3)(4x-1)(5x-2)(x+9)*x-4/x+5
QuotOne
a/a=1
1*x-4/x+5
IdPropMult
Identity Property of Multiplication
x-4/x+5
b
We want to divide the given rational expressions.
x^2-9/x^2+6x+9÷x^2-x-6/x^2+4
To make the division easier, let's factor each of the polynomials.
We will continue with the denominator in the first fraction. We can rewrite the x-term as a product with a factor 2 and the constant as a square. Then we can factor the perfect square trinomial.
x^2+6x+9
SplitIntoFactors
Split into factors
x^2+2(x)(3)+9
WritePow
Write as a power
x^2+2(x)(3)+3^2
FacPosPerfectSquare
a^2+2ab+b^2=(a+b)^2
(x+3)^2
The next polynomial is the numerator in the second fraction. Since it is not a perfect square trinomial, we have to factor it by writing the x-term as two integers that multiply to -6 and add to -1.
x^2-x-6
(x+2)(x-3)
It is not possible to factor x^2+4, since the square of a binomial requires a third term. We can now divide the fractions. When two fractions are divided, it is the same thing as multiplying the first fraction with the reciprocal of the second.
(x+3)(x-3)/(x+3)^2 ÷ (x+2)(x-3)/x^2+4
DivFracByFracDiv
a/b÷c/d=a/b*d/c
(x+3)(x-3)/(x+3)^2 * x^2+4/(x+2)(x-3)
MultFrac
Multiply fractions
(x+3)(x-3)(x^2+4)/(x+3)^2(x+2)(x-3)
PowToProdTwoFac
a^2=a* a
(x+3)(x-3)(x^2+4)/(x+3)(x+3)(x+2)(x-3)
CommutativePropMult
Commutative Property of Multiplication
(x+3)(x-3)(x^2+4)/(x+3)(x-3)(x+3)(x+2)
WriteProdFrac
Write as a product of fractions
(x+3)(x-3)/(x+3)(x-3)*x^2+4/(x+3)(x+2)
QuotOne
a/a=1
1*x^2+4/(x+3)(x+2)
IdPropMult
Identity Property of Multiplication
x^2+4/(x+3)(x+2)
c
To add the whole number and the fraction we want to rewrite the number as a fraction. We want it to have the same denominator as the given fraction. Therefore, 6 is multiplied by x+1x+1.
6+3/x+1
NumberToFrac
a = (x+1)* a/(x+1)
(x+1)*6/x+1+3/x+1
AddFrac
Add fractions
(x+1)6+3/x+1
Distr
Distribute 6
6x+6+3/x+1
AddTerms
Add terms
6x+9/x+1
We could now factor out 3 from the numerator, but it would not make it possible to simplify the fraction further.
d
To subtract the fractions we need to rewrite them with the same denominator. To know what we should expand the first fraction with, let's factor out x from the denominator in the second fraction. It will then be clear what to expand with.
5/x-10/x^2+2x
FactorOut
Factor out x
5/x-10/x(x+2)
ExpandFrac
a/b=a * (x+2)/b * (x+2)
5(x+2)/x(x+2)-10/x(x+2)
SubFrac
Subtract fractions
5(x+2)-10/x(x+2)
Distr
Distribute 5
5x+10-10/x(x+2)
SubTerm
Subtract term
5x/x(x+2)
ReduceFrac
a/b=.a /x./.b /x.
5/x+2
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