(2a-3)(a-1)
5a4 + 3a3 + 2a2
5a3-45a=-2a2+18 5a(a2-9)=-2(a2-9) 5a=-2 a=-2/5
2a2 - 13a + 15 = (2a - 3) (a - 5)
Multiply the first and last coefficients.2*3=6What two factors give you six but when combined give you -5-2 and -3Therefore2x-3)(x-1) will be the factored model.
3a4 - 2a2 + 5a - 10 - 2a4 + 4a2 + 5a - 2 = a4 + 2a2 + 10a - 12 = a4 + (2a - 2)(a + 6)
(2a-3)(a-1)
The other factor is 1.
5a4 + 3a3 + 2a2
(a - 1)(2a - 3)
5a3-45a=-2a2+18 5a(a2-9)=-2(a2-9) 5a=-2 a=-2/5
2a2 - 13a + 15 = (2a - 3) (a - 5)
Multiply the first and last coefficients.2*3=6What two factors give you six but when combined give you -5-2 and -3Therefore2x-3)(x-1) will be the factored model.
7a + 4 = 5a + 7 7a = 5a + 3 2a = 3 a = 3/2
5a
As stated it is just an algebraic expression. However, if we equate it to zero then it can be solved. 2a2 + 5a + 2 = 0 This can be factored (2a + 1)(a + 2) = 0 This means that (2a + 1) = 0 or (a + 2) = 0 Then either 2a = -1 : a = -1/2 or a = -2
13a - 6 + a = 5a + 3 + 3a 13a + a - 5a - 3a = 3 + 6 6a = 9 6a/6 = 9/6 a = 3/2