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Core Connections Algebra 2, 2013 View details

2. Section 3.2

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Exercise 120 Page 159

Practice makes perfect

a The fractions we want to add do not have the same denominator. 1/x+2+3/x^2-4 Let's factor x^2-4 to see if we can find a common denominator.

x^2-4

WritePow

Write as a power

x^2-2^2

FacDiffSquares

a^2-b^2=(a+b)(a-b)

(x+2)(x-2)

This means that we can rewrite the expression as follows. 1/x+2+3/(x+2)(x-2) By multiplying the numerator and denominator of the first fraction by (x-2), we will be able to add the fractions and then simplify.

1/x+2+3/(x+2)(x-2)

ExpandFrac

a/b=a * (x-2)/b * (x-2)

x-2/(x+2)(x-2)+3/(x+2)(x-2)

AddFrac

Add fractions

x-2+3/(x+2)(x-2)

AddTerms

Add terms

x+1/(x+2)(x-2)

b To find how to rewrite the fractions so that they have the same denominator, let's factor the denominators. In the first fraction we can factor out 2.

3/2x+4-x/x^2+4x+4 [0.7em] ⇕ [-0.2em] 3/2(x+2)-x/x^2+4x+4 In the second fraction let's rewrite 4x to 2x+2x and factor.

x^2+4x+4

▼

Factor

WriteSum

Write as a sum

x^2+2x+2x+4

FactorOut

Factor out x

x(x+2)+2x+4

FactorOut

Factor out 2

x(x+2)+2(x+2)

FactorOut

Factor out (x+2)

(x+2)(x+2)

Therefore, the expression can be written as follows. 3/2(x+2)-x/(x+2)(x+2) By multiplying the first numerator and denominator by (x+2) and the second numerator and denominator by 2, the fractions will have the same denominator. Then we can subtract.

3/2(x+2)-x/(x+2)(x+2)

ExpandFrac

a/b=a * (x+2)/b * (x+2)

3(x+2)/2(x+2)(x+2)-x/(x+2)(x+2)

ExpandFrac

a/b=a * 2/b * 2

3(x+2)/2(x+2)(x+2)-2x/2(x+2)(x+2)

SubFrac

Subtract fractions

3(x+2)-2x/2(x+2)(x+2)

Distr

Distribute 3