(x-3)/{(4x-5)(x+1)]
I am lost
2006-11-11 08:40:38 · 4 answers · asked by danni c 1 in Education & Reference ➔ Homework Help
Your denominator can't equal zero because you can't divide by zero. So any time that x is a value that would cause the bottom to be zero, that value needs to be excluded.
Since (4x-5)(0) is 0, and (0)(x+1) is zero, you can break up the bottom two equations and solve for x when the equation equals zero.
4x-5 = 0
4x = 5
x = 5/4
and
x+1 = 0
x = -1
If x = 5/4 or x = -1, the denominator would equal zero and the function would become problematic. So you need to exclude x = 5/4 and x = -1.
2006-11-11 08:47:46 · answer #1 · answered by robtheman 6 · 0⤊ 0⤋
Anything that makes the denominator zero. So, 5/4 and -1
2006-11-11 08:42:41 · answer #2 · answered by just browsin 6 · 0⤊ 0⤋
Expand the double brackets and it becomes (x-3)/4x²-x-5 then work it out from that.
2006-11-11 08:50:22 · answer #3 · answered by Eoin W 1 · 0⤊ 0⤋
just browsin is right
2006-11-11 12:38:20 · answer #4 · answered by alpha 7 · 0⤊ 0⤋