2007-11-03 00:56:58 · 9 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
(3x+1)(2x-5)
2007-11-03 01:04:23 · answer #1 · answered by imalava 2 · 1⤊ 1⤋
6x² - 13x - 5 = 0
The middle term is - 13x
Find the sum of the middle term
Multiply the first term 6 times the last term - 5 equals - 30 and factor
Factors of - 30
1 x - 30
2 x - 15. . .←. .use these factors
3 x - 10
5 x - 6
+ 2 and - 15 satisfy the sum of the middle term.
Insert + 2x and - 15x into the equation
6x² - 13x - 5 = 0
6x² + 2x - 15x - 5 0
Group factor
(6x² + 2x) - (15x - 5) = 0
2x(3x + 1) - 5(3x + 1) = 0
(2x - 5)(3x + 1) = 0
- - - - - - - - s-
2007-11-03 02:03:41 · answer #2 · answered by SAMUEL D 7 · 1⤊ 0⤋
6x^2-13x-5 = (3x-1)(2x+5)
I worked this out because in a quadratic, the form is ax^2+bx+c. The last number in each of the brackets must multiply together to make c. 1 and 5 is the only option here. The first number in each of the brackets must also multiply together to get a. this could be 6 and 1 or 3 and 2. You then use simple trial and improvement to find the correct combination.
If this function equalled 0, the value of x would be either 1/3 or -2.5
2007-11-03 01:08:43 · answer #3 · answered by Alex 2 · 0⤊ 2⤋
6x² -13x - 5
= 6x² - 15x + 2x - 5
= 2x (1 + 3x) - 5 (1 + 3x)
= (2x - 5)(1+3x)
2007-11-03 01:24:24 · answer #4 · answered by gauravragtah 4 · 0⤊ 0⤋
6x^2 - 13x - 5 = (3x + 1)(2x - 5)
2007-11-03 01:08:06 · answer #5 · answered by Anonymous · 0⤊ 1⤋
(a) incorrect = 5x + 5 + 2x + 6 = 7x + eleven (b) incorrect 6x^2 + 15x + 2x +5 = 6x^2 +17x +5 (a) became the question x^2 -7x +6 , this could factorise to (x-a million)(x-6) (b) = (2x)^2 - 3^2 = (2x-3)(2x+3) (a) appropriate (b) appropriate
2016-11-10 03:27:50 · answer #6 · answered by Anonymous · 0⤊ 0⤋
6(x+5/2)*(x-1/3)
2007-11-03 01:07:01 · answer #7 · answered by artie 4 · 0⤊ 1⤋
6X2-13X-5=6X2+2X-15X-5
2X(3X+1)-5(3X+1)
(3X+1)(2X-5)
2007-11-03 01:16:05 · answer #8 · answered by tiya b. 2 · 0⤊ 0⤋
(3x+1)(2x-5)
2007-11-03 01:08:51 · answer #9 · answered by Dr K.L.Verma 2 · 0⤊ 1⤋