In your first question, you know that the function is a parabola, opening upwards parallel to the y-axis. It will be positive, then, wherever the parabola is above the x-axis. Find the roots:

f(x) = (x + 5)(x – 4)

So the roots are x = -5 and x = 4. If you envision the parabola, then, it will be positive on

(-inf, -5] U [4, +inf)

This can also be done algebraically, but it’s harder. Again, you factor it:

(x + 5)(x – 4) >= 0

Now you have two conjunctions: in order to be positive, the factors must either be both positive or both negative. In the first case:

x + 5 >= 0 intersect x – 4 >= 0

x >= -5 intersect x >= 4

That intersection is x >= 4

In the second case,

x + 5 <= 0 intersect x - 4 <= 0 x <= -5 intersect x <= 4 That intersection is x <= -5 The solution is the union of those two: x <= -5 U x >= 4

which is the same as the geometric answer.

Pick the one you like the best. The rest will work the same way.

if there are 0 or 1 answers, then check the coefficient of x^2-

if the coefficient is positive, then (-infinity, infinity)

if the coefficient is negative, then there is no correct answer

if there is 2 answers, then plug in random values greater than both answers, between the answers, and less than both answers & take the sections where the answer is positive…

f(x) = x^2 + x – 20…. (x-4)(x+5)=0, therefore x = 4 or x = -5

now plug in: @ x=0, f(x) = -20… the section (-4, 5) is negative

@ x = 7, f(x)=36, therefore the section [4, infinity) is positive

@ x = -6, f(x) = 22, therefore the section (- infinity, -5] is positive

– answer: ( – infinity, -5] and [4, infinity) for the first question

for the second: x = -8 and x = 5

looking for negative, so plug in numbers again (I will skip this part)

answer: [ -8, 5]

hope that helped!

http://en.wikipedia.org/wiki/Partial_fraction