Question

x(x-1)(x+2)>0 solve the inequality and the answer needs to be in interval notation and i have no idea what im doing

Answer 1

this sucker changes sign at the zeros of the factors, namely at
\[-2,0,1\]

Answer 2

break the real line up in to four parts
\[(-\infty,-2), (-2,0), (0,1), (1,\infty)\]

Answer 3

since this is a polynomial of degree 3 with postive leading coefficient, the graph will be
negative, positive, negative, positive
in that order

Answer 4

ok you lost me there a little bit how do i fins those numbers?

Answer 5

these numbers \(-2,0,1\) ?

Answer 6

yes are they just the ones that are in the equation?

Answer 7

solve \(x(x-1)(x+2)=0\) in your head

Answer 8

so when you graph \(y=x(x-1)(x+2)\) it will cross the \(x\) axis at \(-2\) and \(0\) and \(1\)
as it crosses the \(x\) axis it goes from being negative (below) to positive (above) or vice versa

Answer 9

so the answer in integral form would be -2< x<0 or x> 1

Answer 10

that is not integral form or even interval form, that is as in inequality
but yes that is right

Answer 11

interval form is \((-2,0)\cup ((1,\infty)\)

Answer 12

ic what does the u stand for?

Answer 13

union

Answer 14

i have an extra parentheses there, that is a typo

Answer 15

ok so in order to find the answer you must find the solutions on the graph correct?

Answer 16

no you can do that stupid "test point" business
but it is a waste of time
if you know that \(x(x-1)(x+2)>0\) is a synonym for "positive" and that positive is a synonym for "above the \(x\) axis"
and if you know what a third degree polynomial looks like
then you can do it in your head

Answer 17

otherwise you could test a point, say \(x=-3\) and see that it is negative
then you know it will be \(-,+,-,+\) on the intervals