Martin-Gay, Beginning Algebra, 5ed 22 33 44 55.

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Martin-Gay, Beginning Algebra, 5ed 22

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10 Example For each pair of polynomials, find the least common multiple. a) 16a and 24b b) 24x 4 y 4 and 6x 6 y 2 c) x 2  4 and x 2  2x  8 Solution a) 16a = 2  2  2  2  a 24b = 2  2  2  3  b The LCM = 2  2  2  2  a  3  b The LCM is 2 4  3  a  b, or 48ab 16a is a factor of the LCM 24b is a factor of the LCM

Martin-Gay, Beginning Algebra, 5ed 11 Example continued b) 24x 4 y 4 = 2  2  2  3  x  x  x  x  y  y  y  y 6x 6 y 2 = 2  3  x  x  x  x  x  x  y  y LCM = 2  2  2  3  x  x  x  x  y  y  y  y  x  x Note that we used the highest power of each factor. The LCM is 24x 6 y 4 c) x 2  4 = (x  2)(x + 2) x 2  2x  8 = (x + 2)(x  4) LCM = (x  2)(x + 2)(x  4) x 2  4 is a factor of the LCM x 2  2x  8 is a factor of the LCM

Martin-Gay, Beginning Algebra, 5ed 12 a) b) c) d) e)