If you factor the equation you can see what the solutions are. Two factor a quadratic of the form ax^2+bx+c, find two values which satisfy two conditions...
jk=ac=-15 and j+k=b=-2 so j and k must be -5 and 3 so the factors are:
(x-5)(x+3)
So the other solution is x=5
The count the number of times the sign changes
that is how many positive roots there are
if you get a number that is ≥2, then count down by 2's ending at 0
sub -x for x and evaluate
count change in sign again
that is how many negative roots there are
if you get a number that is ≥2, then count down by 2's ending at 0
so
-2x^3+3x^2-5x-2=0
-,+,-,-
1 2
2 or 0 positive roots
x to -x
2x^3+3x^2+5x-2=0
+,+,+,-
1
1 negative root
2 or 0 positive roots and 1 negative root
D is answer
