2007-05-15 03:50:45 · 6 answers · asked by llaattiinnlloovvee 1 in Science & Mathematics ➔ Mathematics
24x^2 - 4x - 48
= 4 (6x^2 - x - 12)
= 4 (6x^2 - 8x + 9x - 12)
= 4 [2x(3x - 4) + 3(3x - 4)]
= 4(2x+3)(3x-4)
2007-05-15 03:54:44 · answer #1 · answered by wangsacl 4 · 0⤊ 1⤋
1. Divide through to make it simpler. In your case, divide through by four to get:
8x^2-x-12
2. Check to see if you can factor it easily (i.e., you can see how to factor it. This takes a lot of practice). If not, go on to step 3.
3. Use the quadratic formula. Here's a link (scroll down a bit): http://en.wikipedia.org/wiki/Quadratic_equation. In your case, a=8, b= -1 and c= -12. "Plug and chug," as my old math teacher would say.
4. You'll get two answers, in your case both radicals. These are your roots. If you want to factor them, simply subtract the number from x. For example, if your answers are 3 and -2, your factors will be (x-3) and (x+2).
And you're done!
2007-05-15 10:59:55 · answer #2 · answered by Dark Knight 3 · 1⤊ 0⤋
24x^2-4x-48
4(6x^2 - 2x - 12)
4[2(3x^2 - x - 6)]
8(3x^2 - x - 6)
x = -(-1) + - root(6)^2 - 4(3)(-6) / 2(3)
x = 1 + - root36 + 72 / 6
x = 1 + - root108 / 6
x = 1 + - root4 * root27 / 6
x = 1 + - 2*root 27 / 6
x1 = 1 + 2*root 27 / 6 or x2 = 1 - 2*root27 / 6
x1 =1.9(approximately) x2 = -1.6(approximately)
2007-05-15 11:01:02 · answer #3 · answered by md_kath 1 · 0⤊ 1⤋
24x^2-4x-48
(6x + 8)(4x - 6)
2007-05-15 11:00:01 · answer #4 · answered by Robert L 7 · 0⤊ 1⤋
4(6x^2-x-8)
4(3x-4)(x+2)
2007-05-15 10:53:48 · answer #5 · answered by txmama423 3 · 0⤊ 1⤋
6x^2-x-12=0
6x^2-9x+8x-12=0
3x(2x-3)+4(2x-3)=0
(3x+4)(2x-3)
2007-05-15 11:07:00 · answer #6 · answered by dumblum 2 · 0⤊ 1⤋