x^3+x^2+64x+640 find the solution set

Question

x^3+x^2+64x+64>0 find the solution set

zehanz · Answer

First, try to find the solutions of $$x^3+x^2+64x+64=0$$Do you know how to begin?

anonymous · Answer

nope

zehanz · Answer

Try a few "simple" numbers, like 1, -1, 2 or -2...

zehanz · Answer

This is not as dumb as you may think: solving 3rd degree equations is very difficult, so if you have to do it, there have to be simple solutions that you can see without calculations!
One of the four numbers I mentioned is a solution.

anonymous · Answer

1 and 2 are solutions

zehanz · Answer

$$2^3+2^2+64*2+64 = 8+4+128+64 \neq0$$Same for 1 :(

anonymous · Answer

the only one that equals 0 is -1

anonymous · Answer

so is the answer -1

zehanz · Answer

OK, that's right. 
Now you should try to factorize $$x^3+x^2+64x+64$$I mean: $$x^3+x^2+64x+64 = (x+1)(x^2+...)=0$$Can you figure out how?

anonymous · Answer

(x+1)(x^2+64) now what do i do

zehanz · Answer

If you realize what values x^2+64 can take, you'll see that it never can be 0.
After all, x^2 >= 0 for all x, so adding another surely 64 makes it positive.
This means that -1 is indeed the only solution of $$x^3+x^2+64x+64=0$$You now have two areas to investigate: the numbers smaller than -1 and the numbers larger than -1.
So which set of numbers will make$$x^3+x^2+64x+64>0$$

anonymous · Answer

so the solution set is $$xlx \neq0 $$

zehanz · Answer

Try to see it this way: you've got this crazy number, call it P, $$P=x^3+x^2+64x+64$$P can take on all kinds of values, depending on the x you put in. Some values of P are positive, some are negative and it happens only once to be 0, that is when x = -1. You are looking for x that make P positive. The are only two possibilities for this: either every x > -1 will do the trick, or every x < -1. In this case you can try any number to see what happens. Take for instance x = 0 (easy to calculate P). This gives P > 0. So the solution must be: x > -1. If you try a number left of -1, say -2, you will get P < 0.