a5+b5 =
(a+b)
(a4-a3b+a2b2-ab3+b4)
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A = 3B = 5 ___________ 2 ( 3A + B) = 2 (9 + 5) = 2 (14) = 28
T(x) = ax^2 + bx + cT(0) = -4T(1) = -2T(2) = 6Solution:Since T(0) = -4, then c = -4.So, when x = 1, we have:-2 = a(1)^2 + b(1) - 4-2 = a + b - 4-2 + 4 = a + b - 4 + 42 = a + b2 - a = a - a + b2 - a = bWhen x = 2, we have:6 = a(2)^2 + b(2) - 46 = 4a + 2b - 46 + 4 = 4a + 2b - 4 + 410 = 4a + 2b10/2 = 4a/2 + 2b/25 = 2a + b5 - 2a = 2a - 2a + b5 - 2a = b2 - a = 5 - 2a2 - 2 - a + 2a = 5 - 2 - 2a + 2aa = 32 - a = b2 - 3 = b-1 = bThus, a = 3, b = -1, and c = -4.Check:
LCM(a2b5, a3b3) The LCM of both numbers HAS to have the largest coefficient of both variables. For a, it's a3, and for b it is b5. So the LCM is a3b5.
6m b5