I just can't figure out these 4...Would be very happy if someone could help me out
Solve the equations:
1) (2x^3 -3x^2 -3x+2) / (x+1) < 0

2) (x^3 -2x^2 -3x) > 0

Solve the divisions
3) (2x^3 +9x^2 -7x -4) : (x^2 +2x -3)

4) x^4 : (x^2 - 4)
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2. Originally Posted by pizdac
I just can't figure out these 4...Would be very happy if someone could help me out
Solve the equations:
1) (2x^3 -3x^2 -3x+2) / (x+1) < 0
Do the division

$\displaystyle f(x)=\frac{2x^3-3x^2-3x+2}{x+1}=2x^2-5x+2$.

Now as the coefficient of $\displaystyle x^2$ is positive $\displaystyle f(x)<0$ between the roots
of the quadratic on the right (if these are real). The roots are $\displaystyle 2$ and $\displaystyle 1/2$. So:

$\displaystyle f(x)=\frac{2x^3-3x^2-3x+2}{x+1}<0$, when $\displaystyle x \epsilon (1/2,2)$

RonL

3. Originally Posted by pizdac
2) (x^3 -2x^2 -3x) > 0
The LHS of this inequality is a cubic, it has either three real roots, or but
a single root. Also as $\displaystyle x \rightarrow \infty$ the cubic is positive.

So if there is only a single root the cubic is $\displaystyle >0$ to the right
of the root.

If there are three roots it is $\displaystyle >0$ to the right of the greatest root,
and again between the other two roots.

(Sketch a cubic and you will see why this is so).

As there is no constant term in the cubic we can factor it:

$\displaystyle x(x^2 -2x -3)$,

and so we can find the other roots using the quadratic formula.

RonL

4. Problem 4
$\displaystyle \frac{x^4}{x^2-4}=x^2+4+\frac{16}{x^2-4}$

5. Problem 2)
$\displaystyle x^3-2x^2-3x>0$
Factor,
$\displaystyle x(x^2-2x-3)$
Once more,
$\displaystyle x(x-3)(x+1)$.
Now find the zero,
$\displaystyle x=-1,0,3$
Now observe each interval.
$\displaystyle x<-1$
$\displaystyle -1<x<0$
$\displaystyle 0<x<3$
$\displaystyle 3<x$
Just pick points and see if the polynomial is positive or negative.
Doing that we find that
$\displaystyle x<-1$ This is Negative (Wrong)
$\displaystyle -1<x<0$ This is Positive (Correct)
$\displaystyle 0<x<3$ This is Negative (Wrong)
$\displaystyle 3<x$ This is Positive (Correct)
$\displaystyle -1<x<0 \mbox{ or } x>3$
Or another way of writing $\displaystyle x\in (-1,0)\cup (3,\infty)$.