You have gained experience working with rational functions so far this chapter. In this lesson, you will continue simplifying rational expressions by factoring.
To simplify a rational expression means to reduce the fraction into its lowest terms.
To do this, you will need to remember a property about multiplication.
For all real values a,b, and b≠0,abb=a.
Example: Simplify 4x−22x2+x−1.
Both the numerator and denominator can be factored using methods learned in Chapter 9.
The expression (2x−1) appears in both the numerator and denominator and can be canceled. The expression becomes:
Example 1: Simplify x2−2x+18x−8.
Solution: Factor both pieces of the rational expression and reduce.
Finding Excluded Values of Rational Expressions
As stated in Lesson 2 of this chapter, excluded values are also called points of discontinuity. These are the values that make the denominator equal to zero and are not part of the domain.
Example 2: Find the excluded values of 2x+1x2−x−6.
Solution: Factor the denominator of the rational expression.
Find the value that makes each factor equal zero.
These are excluded values of the domain of the rational expression.
Real-Life Rational Expressions
The gravitational force between two objects in given by the formulaF=G(m1m2)(d2). The gravitation constant is given by G=6.67×10−11(N⋅m2/kg2). The force of attraction between the Earth and the Moon is F=2.0×1020N (with masses of m1=5.97×1024kg for the Earth and m2=7.36×1022kg for the Moon).
What is the distance between the Earth and the Moon?
Let's start with the Law of Gravitation formula.Now plug in the known values.Multiply the masses together.Cancel thekg2units.Multiply the numbers in the numerator.Multiply both sides byd2.Cancel common factors.Simplify.Divide both sides by2.0×1020N.Simplify.Take the square root of both sides.F=Gm1m2d22.0×1020N=6.67×10−11N⋅m2kg2.(5.97×1024kg)(7.36×1022kg)d22.0×1020N=6.67×10−11N⋅m2kg2.4.39×1047kg2d22.0×1020N=6.67×10−11N⋅m2kg2⋅4.39×1047kg2d22.0×1020N2.93×1037d2N⋅m22.0×1020N⋅d2=2.93×1037d2⋅d2⋅N⋅m22.0×1020N⋅d2=2.93×1037d2⋅d2⋅N⋅m22.0×1020N⋅d2=2.93×1037N⋅m2d2=2.93×10372.0×1020N⋅m2Nd2=1.465×1017m2d=3.84×108m
Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both.
Find the excluded values for each rational expression.
In an electrical circuit with resistors placed in parallel, the reciprocal of the total resistance is equal to the sum of the reciprocals of each resistance: 1Rc=1R1+1R2. If R1=25Ω and the total resistance is Rc=10Ω, what is the resistance R2?
Suppose that two objects attract each other with a gravitational force of 20 Newtons. If the distance between the two objects is doubled, what is the new force of attraction between the two objects?
Suppose that two objects attract each other with a gravitational force of 36 Newtons. If the mass of both objects was doubled, and if the distance between the objects was doubled, then what would be the new force of attraction between the two objects?
A sphere with radius r has a volume of 43πr3 and a surface area of 4πr2. Find the ratio of the surface area to the volume of the sphere.
The side of a cube is increased by a factor of two. Find the ratio of the old volume to the new volume.
The radius of a sphere is decreased by four units. Find the ratio of the old volume to the new volume.
Simplify (4b2+b+7b3)+(5b2−6b4+b3). Write the answer in standard form.
State the Zero Product Property.
Why can’t the Zero Product Property be used in this situation: (5x+1)(x−4)=2?
Shelly earns $4.85 an hour plus $15 in tips. Graph her possible total earnings for one day of work.
Multiply and simplify: (−4x2+8x−1)(−7x2+6x+8).
A rectangle’s perimeter is 65 yards. The length is 7 more yards than its width. What dimensions would give the largest area?