2006-11-05 11:40:52 · 9 answers · asked by HW Help 1 in Science & Mathematics ➔ Mathematics
(3x-2)(x-3)
2006-11-05 11:48:36 · answer #1 · answered by ... 6 · 0⤊ 0⤋
I am going to use the factorisation method of splitting the middle term. In order to do it you must first multiply the coefficient of the first term with the last term.
3*6 = 18
Now, we need two factors of 18 that add up to - 11. Obviously, both must be negative. Let's consider the possibilities:
-1*-18
-2*-9
I'll stop here as we've found the numbers we need
-2-9 = -11
-2*-9 = 18
Now, let's get on with the factorisation:
3x^2 -11x+6 = 3x^2 - 9x - 2x + 6
= 3x(x - 3) - 2(x - 3)
= (3x - 2)(x - 3)
2006-11-05 12:15:07 · answer #2 · answered by Akilesh - Internet Undertaker 7 · 0⤊ 1⤋
3x^2-11x+6=0
(3x^2-9x)-(2x+6)=0
3x(x-3)-2(x-3)=0,therefore the factor of3x^2-11x+6=
(3x-2)(x-3)
2006-11-05 11:51:45 · answer #3 · answered by Anonymous · 0⤊ 0⤋
3x^2 - 11x + 6
x = (-b ± sqrt(b^2 - 4ac))/(2a)
x = (-(-11) ± sqrt((-11)^2 - 4(3)(6)))/(2(3))
x = (11 ± sqrt(121 - 72))/6
x = (11 ± sqrt(49))/6
x = (11 ± 7)/6
x = (18/6) or (4/6)
x = 3 or (2/3)
ANS : (3x - 2)(x - 3)
2006-11-05 11:48:36 · answer #4 · answered by Sherman81 6 · 0⤊ 0⤋
the answer is (3x-2)(x-3)
2006-11-05 11:46:48 · answer #5 · answered by reem h 2 · 0⤊ 0⤋
(3x-2)(x-3)
2006-11-05 11:43:35 · answer #6 · answered by Anonymous · 0⤊ 0⤋
(3x-2)(x-3)
2006-11-05 11:46:47 · answer #7 · answered by Anonymous · 0⤊ 0⤋
3x^2-11x+6=
65-3=2=
567$$$888****
niner...
i'm retarded.
2006-11-05 11:47:40 · answer #8 · answered by Js_5 5 · 0⤊ 0⤋
(x-3)*(3x-2)
Doug
2006-11-05 11:48:10 · answer #9 · answered by doug_donaghue 7 · 0⤊ 1⤋