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Mathematics 18 Online
OpenStudy (anonymous):

1. [8.01] Identify the GCF for 12a4b3 − 18a3b2 + 24a2b. (3 points) 6a2b2 12a2b 12ab 6a2b 2. [8.01] Factor completely: −9a2 + 18a − 27 (4 points) −1(9a2 − 18a + 27) −9a(a2 − 2a + 3) −9(a2 + 2a − 3) −9(a2 − 2a + 3) 3. [8.01] Factor completely: 8m3n2 − 6m2n3 (4 points) 2mn(4m2n − 3mn2) 2m3n3(4m − 3n) m2n2(8m − 6n) 2m2n2(4m − 3n) 4. [8.01] Factor completely: 15a4b − 10a3b2 + 5a2b3 (4 points) 5ab(3a2 − 2ab + b2) 5ab2(3a2 − 2ab + b2) 5a2b(3a2 + 2ab + b2) 5a2b(3a2 − 2ab + b2)

OpenStudy (callisto):

First take a look at the first question 12a4b3 − 18a3b2 + 24a2b you can see that 12, 18, 24 are 6 multiples, in another words, 6 is the common factor of them The 3 terms also contain a^2 and b So the GCF = 6a^2b

OpenStudy (callisto):

For factorization problem, first take out the common factor −9a2 + 18a − 27 = -9 (a^2 -2a +3) Consider a^2 -2a +3 a^2 -2a -3 = a^2 -3a + a -3 Take out the common factor again a(a-3) +(a-3) group the like terms together (a+1)(a-3) Therefore, −9a2 + 18a − 27 = -9 (a^2 -2a +3) = -9(a^2 -3a + a -3) = -9[a(a-3) +(a-3)] = -9 (a+1)(a-3) = 9 (3-a)(a+1) Nah.. I'm pretty sure that there is a problem in the answers - they are not factored completely ...

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