Finding the LCD (Least Common Denominator)

Math A-100 IUSB Sec 7.4 Detailed notes – Week 12 Finding the LCD (Least Common Denominator Finding the LCD If you are asked for just the LCM, 1) pretend that the polynomials they give you are the bottoms of polynomial fractions, 2) factor both denominators to find out what is missing from each denominator. 3) Change the fraction to add in the missing part, both on top & bottom (just like with regular fractions 4) When you put all the parts together, you have the LCD. NOTE: Once the denominators are the same  this is the LCD (Least Common Denominator) Example 1: Find the LCD 7___ & 2x + 4 24x 6 y 2 16x 3 y 3 Different Denominators  3 different “ things ” : # ’ s, x ’ s, & y ’ s Look at each individually to see what is missing # ’ s: 24 & 16  both go into 48 x ’ s: x 6 & x 3  both go into x 6 y ’ s: y 2 & y 3  both go into y 3 Put the parts back together: 48 x 6 y 3  LCD Example 2: Find the LCD 3 & 2x + 4 x x 2 + 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 becomes 3 (x + 2) & 2x + 4 bottoms are the same now x x(x + 2) x (x + 2) x(x + 2) bottoms are the same now  LCM = 2(x + 2)

Example 3: Find the LCD x & 3 x 2 – 3x + 2 x 2 - 5x + 4 Different Denominators  factor the denominators to find the LCM x & 3 becomes x & 3______ x 2 – 3x + 2 x 2 - 5x + 4 (x -1) (x – 2) (x – 1) (x – 4) 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) & 3 (x - 2) _ (x -1)(x – 2) (x – 4) (x – 1)(x – 4) (x – 2) LCM = (x – 1)(x – 2)(x – 4) Example 4: Find the LCD x + 3 & x + 2 x 2 - 2x - 8 x 2 - 4 Different Denominators  factor the denominators to find the LCM x + 3 & x + 2 becomes x + 3 & x + 2 x 2 - 2x - 8 x 2 - 4 ( x - 4)( x +2) (x – 2)( x + 2) what is “ missing ” ?: the left fraction bottom is missing the (x - 2) part the right fraction is missing the (x – 4) part Change both fractions to include the “ missing ” part: (x + 3) (x – 2) & (x + 2) (x – 4) ___ (x -4)(x +2) (x – 2) (x – 2)(x + 2) (x – 4) LCM = (x + 2)(x – 2)(x – 4)

Special Cases “ a – b & b – a ” Recall from Section 7.3, we had the situations where the denominators are subtractions in the opposite order, such as y – 1 & 1 – y. We can find the LCD for these too by factoring out -1, which switched the order for one denominator Example 1: Find the LCD 2y 2 + 5y + 3 y - 1 1 – y Factor out -1 for one of the denominators: y – 1  -1(1 – y) LCM = 1 – y NOTE: you don ’ t have to worry about the -1 when finding the LCM . However, you do have to divide it into all parts of the top if asked to add or subtract, like in Section 7.3.