Core Connections Algebra 2, 2013
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Core Connections Algebra 2, 2013 View details
2. Section 3.2
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Exercise 105 Page 153

Practice makes perfect
a Let's look at the expression we are given.
(x-4)^3(2x-1)/(2x-1)(x-4)^2First, we can rearrange the factors to create a Giant One (a form of the number one) and simplify. Let's try it!
(x-4)^3(2x-1)/(2x-1)(x-4)^2
(x-4)^3(2x-1)/(x-4)^2(2x-1)
(x-4)^3/(x-4)^2*2x-1/2x-1
(x-4)^3/(x-4)^2 * 1
(x-4)^3/(x-4)^2
We can simplify this further using the Quotient of Powers Property.
(x-4)^3/(x-4)^2
(x-4)^(3-2)
(x-4)^1
x-4
b We are again given a rational expression.
7m^2-22m+3/3m^2-7m-6 To simplify, let's factor the numerator and denominator.
7m^2-22m+3
â–Ľ
Factor
7m^2-m-21m+3
m(7m-1)-21m+3
m(7m-1)-3(7m-1)
(m-3)(7m-1)
The numerator can be factored to (m-3)(7m-1). Now we factor the denominator.
3m^2-7m-6
â–Ľ
Factor
3m^2-9m+2m-6
3m(m-3)+2m-6
3m(m-3)+2(m-3)
(3m+2)(m-3)
Using the factored forms, we can now simplify the expression.
7m^2-22m+3/3m^2-7m-6
(m-3)(7m-1)/(3m+2)(m-3)
(7m-1)(m-3)/(3m+2)(m-3)
7m-1/3m+2* m-3/m-3
7m-1/3m+2* 1
7m-1/3m+2
c Let's look at the given fraction.
(z+2)^9(4z-1)^7/(z+2)^(10)(4z-1)^5 To simplify this expression we can use that a power divided by another power with the same base can be simplified by subtracting the exponents.
(z+2)^9(4z-1)^7/(z+2)^(10)(4z-1)^5
(z+2)^9/(z+2)^(10)*(4z-1)^7/(4z-1)^5
(z+2)^(-1)(4z-1)^2
1/(z+2)*(4z-1)^2
(4z-1)^2/z+2
d The numerator and denominator of this equation are not factored completely.
(x+2)(x^2-6x+9)/(x-3)(x^2-4) Let's factor the all of the expressions in the numerator and denominator before simplifying.
x^2-6x+9
(x-3)^2
(x-3)(x-3)
Now let's factor the expression from the denominator.
x^2-4
x^2-2^2
(x+2)(x-2)
The second factor in the denominator can be factored to (x+2)(x-2). We can use these factored forms to simplify the original expression.
(x+2)(x^2-6x+9)/(x-3)(x^2-4)
(x+2)(x-3)(x-3)/(x-3)(x+2)(x-2)
(x+2)(x-3)(x-3)/(x+2)(x-3)(x-2)
(x+2)(x-3)/(x+2)(x-3)*x-3/x-2
1*x-3/x-2
x-3/x-2