• January 22nd 2006, 09:13 AM
pizdac
I just can't figure out these 4...Would be very happy if someone could help me out :o
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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• January 22nd 2006, 09:54 AM
CaptainBlack
Quote:

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

Do the division

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

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

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

RonL
• January 22nd 2006, 10:15 AM
CaptainBlack
Quote:

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 $x \rightarrow \infty$ the cubic is positive.

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

If there are three roots it is $>0$ 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:

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

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

RonL
• January 22nd 2006, 10:52 AM
ThePerfectHacker
Problem 4
$\frac{x^4}{x^2-4}=x^2+4+\frac{16}{x^2-4}$
• January 22nd 2006, 11:00 AM
ThePerfectHacker
Problem 2)
$x^3-2x^2-3x>0$
Factor,
$x(x^2-2x-3)$
Once more,
$x(x-3)(x+1)$.
Now find the zero,
$x=-1,0,3$
Now observe each interval.
$x<-1$
$-1
$0
$3
Just pick points and see if the polynomial is positive or negative.
Doing that we find that
$x<-1$ This is Negative (Wrong)
$-1 This is Positive (Correct)
$0 This is Negative (Wrong)
$3 This is Positive (Correct)
$-13$
Or another way of writing $x\in (-1,0)\cup (3,\infty)$.