Core Connections Algebra 2, 2013
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Core Connections Algebra 2, 2013 View details
Chapter Closure

Exercise 131 Page 251

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>0This means we only have to consider the denominator in the first fraction. If we set it equal to 0 and solve for x, we can determine the values of x where the expression is undefined. x+3=0 &⇔ x=-3 As we can see, x cannot equal - 3. To add the fractions, we first have to give them a common denominator.
5x/x+3 + 3+x/x^2+9
5x( x^2+9 )/(x+3)( x^2+9 ) + 3+x/x^2+9
5x( x^2+9 )/(x+3)( x^2+9 ) + (3+x)(x+3)/( x^2+9 ) (x+3)
5x( x^2+9 )+(3+x)(x+3)/(x+3)( x^2+9 )
Simplify numerator
5x^3+45x+(3+x)(x+3)/(x+3)( x^2+9 )
5x^3+45x+(x+3)(x+3)/(x+3)( x^2+9 )
5x^3+45x+(x+3)^2/(x+3)( x^2+9 )
5x^3+45x+x^2+6x+9/(x+3)( x^2+9 )
5x^3+51x+x^2+9/(x+3)( x^2+9 )
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.
x/x-1-1
x/x-1-x-1/x-1
x-(x-1)/x-1
x-x+1/x-1
1/x-1
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/0So far we know that x≠ -2. Next, we will set the denominator in the first fraction equal to 0 and solve for x.
x^2-4x=0
Solve for x
x* x-4* x=0
x(x-4)=0
lx=0 x-4=0
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
(x^2+5x+6)4x/(x^2-4x)(x+2)
(x^2+5x+6)4x/(x* x-4* x)(x+2)
(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
x^2+5x+(2)(3)
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)
(x+3)4x/x(x-4)
(x+3)4/x-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 The left-hand side matches the look of a perfect square trinomial.
x^2-4x+4=0
Factor
x^2-4x+2^2=0
x^2-2(x)(2)+2^2=0
(x-2)^2=0
Solve for x
x-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
x^2-2x/(x-2)^2* x-2/4x^2
(x^2-2x)(x-2)/(x-2)^2 4x^2
Factor
(x* x-2* x)(x-2)/(x-2)^2 4x^2
x(x-2)(x-2)/(x-2)^2 4x^2
x(x-2)^2/(x-2)^2 4x^2
(x-2)^2/(x-2)^2 4x
1/4x
Having simplified the expression, we also see that x cannot equal 0.