x^3-9x^2+14x/x^2-8x
2007-04-16 09:53:44 · 4 answers · asked by crazy 2 1 in Science & Mathematics ➔ Mathematics
x^2-8x can´t be zero
so
x(x-8)=0 x=0 and x=8 must be excluded
2007-04-16 09:57:43 · answer #1 · answered by santmann2002 7 · 0⤊ 0⤋
You have to exclude 0: in the term 14x/x^2, it would be incorrect to divide by 0.
Actually, in response to the above answer, if you're actually dividing 14x by (x^2 - 8x), she's right. You need to be careful in how you write your equations.
2007-04-16 09:59:42 · answer #2 · answered by Sci Fi Insomniac 6 · 0⤊ 0⤋
(6x + 8) / (2x + 10) bear in mind that dividing with the aid of 0 creates an undefined fee. for this reason, the denominator, 2x + 10, won't be able to equivalent 0 2x + 10 ? 0 2x ? -10 x ? -5 -5 is excluded from the area.
2016-12-16 07:39:53 · answer #3 · answered by Anonymous · 0⤊ 0⤋
you're looking for what would make the denominator zero so you're only concerned with x^2-8x.
factor and set equal to zero
x(x-8)=0
x=0 or x=8 <------the 2 values that must be excluded from the domain
2007-04-16 10:02:29 · answer #4 · answered by Shawn F 2 · 0⤊ 0⤋