6z to the 2nd power - 5z - 4 factor each trinomial, if possible. if trinomial cannot be factored using intergers, write prime
2007-04-23 01:21:54 · 3 answers · asked by Anonymous in Education & Reference ➔ Homework Help
6z^2-5z-4
(3z-4)(2z+1)
2007-04-23 01:42:06 · answer #1 · answered by Dave aka Spider Monkey 7 · 1⤊ 1⤋
I'll try to explain how to work these kinds of problems: Let's do number 1: First, you need to find the factors or multiples of -2, and then 6 (aka factors of a and c ). For -2, you get -2 and 1, -1 and 2. Then you fin the multiples of 6: 3 and 2, 6 and 1, 1 and 6, and 3 and 2. Now you will plug those values into the equation as follows: (6x-2)(x +1). Then you multiply -2 and x, and 6x and 1, and add those two expressions together. So you will get -4x. in order for the factor to be correct, the two expressions should add up to bx in the equation, or -x in this case. So 6x-2x... Thats not right. lets try again: (3x-2)(2x+1). -4x+3x=-x. Yeah!!!! That is the answer to number one. 2. (5t-2)(2t+1) 3. (3h-4)(3h+2)
2016-05-17 04:41:38 · answer #2 · answered by ? 3 · 0⤊ 0⤋
man!!! this is a binomial not a trinomial
anyway factors.. hmm okay here goes
6z^2-5z-4
=6z^2 + 3z - 8z - 4
=3z(2z+1) - 4(2z+1)
=(3z-4)*(2z+1)
hence if given that 6z^2-5z-4 = 0 then
(3z-4)*(2z+1)=0
=> either (3z-4) = 0
or (2z+1) = 0
so z = 4/3
or z = -1/2
2007-04-23 01:32:55 · answer #3 · answered by SuNiL 3 · 0⤊ 1⤋