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Module: 0 Part 4: Rational Expressions

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1 Module: 0 Part 4: Rational Expressions
Obj: Simplify rational expressions, complex fractions Evi: Students will solve rational expressions

2 Rational Expressions The set of real numbers for which an algebraic expression is defined is the domain of the expression. π‘₯+2 π‘₯βˆ’3 This is a rational expression the domain is all real numbers expect x=3. If x=3, the bottom is zero and then Channing Tatum dies… we don’t want that.

3 Simplifying a Rational Expression
Step 1: Factor Completely Step 2: Divide out common factors Step 3: State answer and restrictions π‘₯ 2 +4π‘₯βˆ’12 3π‘₯βˆ’6 (π‘₯+6)(π‘₯βˆ’2) 3(π‘₯βˆ’2) π‘₯ π‘₯β‰ 2

4 Another Example 12+π‘₯βˆ’ π‘₯ 2 2 π‘₯ 2 βˆ’9π‘₯+4 βˆ’ π‘₯+3 2π‘₯βˆ’ π‘₯=4, Β± 1 2

5 Multiplying Rational Expressions
Step 1: Factor both of them Step 2: Cancel out common factors Step 3: Figure out domain 2 π‘₯ 2 +π‘₯βˆ’6 π‘₯ 2 +4π‘₯βˆ’5 βˆ— π‘₯ 3 βˆ’3 π‘₯ 2 +2π‘₯ 4 π‘₯ 2 βˆ’6π‘₯ (2+3)(π‘₯+2) (π‘₯+5)(π‘₯βˆ’1) βˆ— π‘₯( π‘₯ 2 βˆ’3π‘₯+2) 2π‘₯(2π‘₯βˆ’3) (2π‘₯βˆ’3)(π‘₯+2) (π‘₯+5)(π‘₯βˆ’1) βˆ— π‘₯(π‘₯βˆ’2)(π‘₯βˆ’1) 2π‘₯(2π‘₯βˆ’3) (π‘₯+2) (π‘₯+5) βˆ— (π‘₯βˆ’2) 2 (π‘₯+2)(π‘₯βˆ’2) 2(π‘₯+5) π‘₯β‰ 0, 1, βˆ’5, 3 2

6 Dividing Rational Expressions
Same as multiplying but you have to switch so that it becomes a multiply by the reciprocal

7 Adding and Subtracting
Step 1: Get common denominator Step 2: Simplify π‘₯ π‘₯βˆ’3 βˆ’ 2 3π‘₯+4 3π‘₯+4 π‘₯βˆ’ π‘₯βˆ’3 2 (π‘₯βˆ’3)(3π‘₯+4) 3 π‘₯ 2 +2π‘₯+6 (π‘₯βˆ’3)(3π‘₯+4)

8 Another Example 3 π‘₯βˆ’1 βˆ’ 2 π‘₯ + π‘₯+3 π‘₯ 2 βˆ’1 2( π‘₯ 2 +3π‘₯+1) π‘₯(π‘₯+1)(π‘₯βˆ’1)

9 Complex Fractions 2 π‘₯ βˆ’3 1βˆ’ 1 π‘₯βˆ’1 2 π‘₯ βˆ’3 2 π‘₯ βˆ’ 3π‘₯ π‘₯ 2βˆ’3π‘₯ π‘₯
Step 1: Separate fractions Step 2: Find Common Denominator Step 3: Simplify Step 4: Multiple by reciprocal Step 5: Simplify 2 π‘₯ βˆ’3 1βˆ’ 1 π‘₯βˆ’1 2 π‘₯ βˆ’ π‘₯ βˆ’ 3π‘₯ π‘₯ βˆ’3π‘₯ π‘₯ 1βˆ’ 1 π‘₯βˆ’ π‘₯βˆ’1βˆ’1 π‘₯βˆ’ π‘₯βˆ’2 π‘₯βˆ’1 2βˆ’3π‘₯ π‘₯ π‘₯βˆ’2 π‘₯βˆ’ βˆ’3π‘₯ π‘₯ Γ· π‘₯βˆ’2 π‘₯βˆ’ βˆ’3π‘₯ π‘₯ βˆ— π‘₯βˆ’1 π‘₯βˆ’2 (2βˆ’3π‘₯)(π‘₯βˆ’1) π‘₯(π‘₯βˆ’2)


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