factor the equation into two binomials
2006-09-10 10:23:55 · 8 answers · asked by maroon_passport 1 in Science & Mathematics ➔ Mathematics
12x^2 - 19x + 5
first, multiply 12 and 5 to give you 60. you use this method to factor quadratics without a x^2 coefficient of 1
x^2 - 19x + 60
factor
(x - 4)(x - 15)
put the 12 back in...
(12x - 4)(12x - 15)
factor AGAIN and drop the factor...(pull out a 4 from the 1st term and a 3 from the second and get rid of them)
(3x - 1)(4x - 5)
2006-09-10 11:18:42 · answer #1 · answered by Giovanni McAdoo 4 · 0⤊ 1⤋
Expression
12x^2-19x+5
Result
(3x - 1)(4x - 5)
Good Luck.
2006-09-10 10:35:43 · answer #2 · answered by sweetie 5 · 0⤊ 1⤋
12x^2-19x+5 .. it's in the same form as ax^2+bx+c : where a =12 b = -19 and c =5
delta = b^2-4ac =19^2-4(12*5)=121 >0 then
x1 =(-b-sqr(delta))/2a=(19-sqrt(121))/2*12 = 8/24
x2 = (-b+sqr(delta))/2a=(19+sqrt(121))/2*12 = 30/24
then
12x^2-19x+5 = (x-x1)(x+x2) =(x-8/24)(x+30/24)= (x*24/8-8/24*24/8)(x*24/6+30/24*24/6)=(3x-1)(4x+5)
then 12x^2-19x+5 = (3x-1)(4x+5)
2006-09-10 11:15:12 · answer #3 · answered by Gaztastic 2 · 0⤊ 1⤋
12x^2 - 19x + 5 = (3x - 1)(4x - 5)
2006-09-10 12:17:59 · answer #4 · answered by Sherman81 6 · 0⤊ 1⤋
OK I LIKE THIS
12x^2-19x+5
(3x-1)(4x-5)
CHECK 12X^ -15X-4X+5
COMBINE 12X^-19X+5
2006-09-10 11:18:42 · answer #5 · answered by juliotelehit 2 · 0⤊ 1⤋
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2016-12-15 05:44:53 · answer #6 · answered by wexler 4 · 0⤊ 0⤋
(3x-1)(4x-5)
2006-09-10 10:30:47 · answer #7 · answered by firstlennsman 1 · 0⤊ 1⤋
(3x-1)(4x-5)
2006-09-10 10:25:49 · answer #8 · answered by spoof ♫♪ 7 · 0⤊ 1⤋