They are numbers of the form 2a*3b where a is 0, 1, ... , 6 and b = 0, 1 or 2.
(3b - 1)(a - b)
3b
Usually, the easiest way to do this is to find a pair of numbers whose sum is equal to the coefficient of the first term, and whose product is equal to the product of the coefficients of the first and last terms: a + b = -3 a * b = -10 take all possible pairs of numbers that multiply to make negative ten: 1 * -10 = -10 -1 * 10 = -10 2 * -5 = -10 -2 * 5 = -10 and then find which one of those pairs also adds up to negative three: 1 + (-10) = -9 -1 + 10 = 9 2 + (-5) = -3 -2 + 5 = 3 So 2 and -5 are the coefficients we want. Now take our original equation: 10a2 - 3a - 1 and separate the middle term into two terms using those coefficients: 10a2 - 5a + 2a - 1 You can then factor a common term out of the first pair and last pair of terms: 5a(2a - 1) + 1(2a - 1) and then group your coefficients: (5a + 1)(2a - 1) Giving you the answer.
ex: 2x2 + 6bx + 5x + 15b Group 1st 2 terms and last 2 terms. Factor separately. = (2x2 + 6bx) + (5x + 15b) Factor using gcf = 2x(x + 3b) + 5(x + 3b) notice the parenthesis are the same = (2x + 5)(x + 3b)
(6ab + 9b)/(2a + 3) = 3b(2a + 3)/(2a + 3) = 3b
2a x 3b = 6ab
2a x 3b = 6ab
0.3333
3a + (3b - 2a) + 5b =3a + 3b - 2a + 5b =(3a - 2a) + (3b + 5b) =a + 8b
2A plus 3B times 2A - 3B = 4A2 - 9B2; this is an example of the general formula (a + b)(a - b) = a2 - b2.
(2a + b)(a + 3b)
12a + 6b + 6c = 6(2a + b + c)
6ab-3b factorize = 3
6ab / 3b = 2a so b / b cancels each other out 6 / 3 = 2 / 1 Example: a = 4 and b = 5 6ab = 6 x 4 x 5 = 120 3b = 3 x 5 = 15 120 / 15 = 24 / 3 = 8 = 2a = 2 x 4
1a+3b 2a+4b=3a +7b
11a+7b-(3b-2a)