Math A-100 IUSB Sec 7.5 Detailed notes – Week 13 Adding & Subtracting Rational Polynomials with different denominators Different Denominators Just like with regular fractions, if the denominators are not the same, you must find a common denominator & change the numerators accordingly before you add or subtract. Recall from Section 7.4, to find the common denominator: 1) factor both denominators to find out what is missing from each denominator 2) change each fraction (top & bottom) to add in the “missing” parts Once both fractions have the same denominator, add or subtract like normal. NOTE: Once the denominators are the same  this is the LCD (Least Common Denominator) Example 1: Add & Simplify 3 + 2x + 4 x x2 + 2x Different Denominators  factor the denominators to find the LCM 3 + 2x + 4 x x(x + 2) Look to see what is “missing”  the left fraction bottom is missing the (x + 2) part, so put is in 3 + 2x + 4 x x(x + 2) becomes 3(x + 2) + 2x + 4 x(x + 2) x(x + 2) bottoms are the same now bottoms are the same now  LCM = 2(x + 2) 3(x + 2) + 2x + 4 x(x + 2) x(x + 2) becomes 3(x + 2) + 2x + 4 x(x + 2) Distribute & Clean it up: 3(x + 2) + 2x + 4 x(x + 2) Factor & simplify: 5(x + 2)  x(x + 2) Example 2: Subtract & Simplify x _ x2 – 3x + 2 becomes 3x + 6 + 2x + 4  5x + 10 x(x + 2) x(x + 2) 5(x + 2)  x(x + 2) 3 x2 - 5x + 4 5 x Different Denominators  factor the denominators to find the LCM x _ 3 becomes x _ 3______ 2 x – 3x + 2 x - 5x + 4 (x -1)(x – 2) (x – 1)(x – 4) 2 what is “missing”?: the left fraction bottom is missing the (x - 4) part the right fraction is missing the (x – 2) part Change both fractions to include the “missing” part: x(x – 4) (x -1)(x – 2)(x – 4) _ 3(x - 2) _ (x – 1)(x – 4)(x – 2) NOTE: LCM = (x – 1)(x – 2)(x – 4) Make one fraction: x(x – 4) - 3(x - 2) (x -1)(x – 4)(x – 2) _ Distribute & Clean it up: x2 – 4x - 3x + 6 (x -1)(x – 4)(x – 2)  factor & cancel: x2 – 7x + 6 (x -1)(x – 4)(x – 2) x2 – 7x + 6__ becomes (x – 6)(x -1)__ (x -1)(x – 4)(x – 2) (x -1)(x – 4)(x – 2) (x – 6) Final Answer: (x – 4)(x – 2) NOTE: as mentioned in Sec 65.1 notes, leave your final answer in factored form. Example 3: Subtract & Simplify x+3 x2 - 2x - 8 - x+2 x2 - 4 Different Denominators  factor the denominators to find the LCM x+3 x2 - 2x - 8 - x+2 x2 - 4 becomes x+3 (x - 4)(x +2) what is “missing”?: the left fraction bottom is missing the (x - 2) part x+2 (x – 2)(x + 2) the right fraction is missing the (x – 4) part Change both fractions to include the “missing” part: (x + 3) (x – 2) (x -4)(x +2) (x – 2) - (x + 2)(x – 4) ___ (x – 2)(x + 2)(x – 4) NOTE: LCM = (x + 2)(x – 2)(x – 4) Make one fraction: (x + 3) (x – 2) - (x + 2)(x – 4) ___ (x -4)(x +2)(x – 2) Distribute & Clean up: 2 2 (x + 1x – 6) - (x -2x – 8)  (x -4)(x +2)(x – 2) x + 1x – 6 - x + 2x + 8)  (x -4)(x +2)(x – 2) 2 Top does not factor  final answer: 2 3x -2______ (x -4)(x +2)(x – 2) 3x -2______ (x -4)(x +2)(x – 2)