how do i solve for x
3x (x+2)(x-2)=0
how do i divide this equation.
(3x – 2)(x - 4) – (x - 4)(6 – 5x) / (4 – x)(8x – 1)
2007-10-30 05:40:19 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
3x(x+2)(x-2)=0
x is those values for which each term = 0...
x = 0, -2, +2
(3x – 2)(x - 4) – (x - 4)(6 – 5x) / (4 – x)(8x – 1)
(x - 4)*((3x – 2) - (6 – 5x)) / -(x - 4)(8x – 1)
((6 – 5x) - (3x – 2)) / (8x – 1)
I think that's all you can do.
2007-10-30 05:45:34 · answer #1 · answered by gebobs 6 · 0⤊ 1⤋
3x(x+2)(x-2)=0
If you have a polynomial = 0, you factor the polynomial and set each factor to zero and solve for the variable value making the statement true
Here the factoring is already done, so
x = 0, x = -2, x = 2
[(3x – 2)(x - 4) – (x - 4)(6 – 5x)] / [(4 – x)(8x – 1)]
I assume you meant to include the brackets [ ]
Note that both terms in the numerator have the factor (x - 4) and that the factor (4 - x) in the denominator is the same as -(x - 4), so now we have
-[(3x - 2) - (6 - 5x)]/(8x - 1)
-[3x - 2 - 6 + 5x]/(8x - 1)
-(8x - 8)/(8x - 1)
-8(x - 1)/(8x - 1)
2007-10-30 12:59:17 · answer #2 · answered by kindricko 7 · 0⤊ 0⤋
To solve for X
3x(x+2)(x-2)=0
-Distribute
3x+6x*3x-6x=0
-Add
9x*(-3x)=0
-Multiply
(-18x)=0
-Divide by (-18) on both sides
x=0
Sorry i can't really help with the other one because i am not in Algebra II
2007-10-30 12:47:31 · answer #3 · answered by Kagomepwndu 2 · 0⤊ 1⤋
In your first equation you have three factors with product = 0. That means at least one of them = 0, so solve these three equations for x:
3x=0
x+2=0
x-2=0
2007-10-30 12:46:57 · answer #4 · answered by DWRead 7 · 0⤊ 0⤋
3x squared=9x
(9x+6x)=15x
15x(x-2)=225x-30x
15x+225x-30x=15x +195x=210x
2007-10-30 12:51:35 · answer #5 · answered by ViNi Da PoOh 2 · 0⤊ 1⤋