For both questions below, solve each equation in the real number system.
[1] x^3 + (3x^2/2) + 3x - 2 = 0
[2] 2x^4 + x^3 - 24x^2 + 20x + 16 = 0
NOTE: I do not understand the wording "...solve each equation in the real number system."
$\displaystyle x^3 + (3x^2/2) + 3x - 2 = 0$
By trial and error (or the way I did it was to plot the graph), you'll find that $\displaystyle x = \frac{1}{2}$ is a solution.
Indeed, by polynomial division, we see that:
$\displaystyle x^3 + (3x^2/2) + 3x - 2 = (x - \frac{1}{2})(x^2 + 2x + 4)$ and we know that $\displaystyle x^2 + 2x + 4$ cannot be factorised over the Real field as its determinant is less than 0.
$\displaystyle 2x^4 + x^3 - 24x^2 + 20x + 16 = 0$
Once again, by trial and error (or plot the graph), you'll find that $\displaystyle x = 2$ is a solution.
Factorising this:
$\displaystyle 2x^4 + x^3 - 24x^2 + 20x + 16$
$\displaystyle = (x - 2)(2x^3 + 5x^2 - 14x - 8)$
$\displaystyle = (x - 2)(x + \frac{1}{2})(2x^2 + 4x - 16)$, for neater form:
$\displaystyle = (x - 2)(2x + 1)(x^2 + 2x - 8)$
We know we can't simplify this further as $\displaystyle (x^2 + 2x - 8)$ has a determinant which is less than 0, so this second question has $\displaystyle x = 2$ and $\displaystyle x = -\frac{1}{2}$
In the event you don't know a way to start finding solutions to these (AlvinCY used, presumably logical, guesswork) you can use the "Rational Root Theorem" to come up with some possibilities.
Given $\displaystyle px^n + ax^{n-1} + ... + bx + q = 0$, any rational roots of this polynomial equation will be of the form: $\displaystyle \pm \frac{\text{factor of q}}{\text{factor of p}}$. There may, of course, be other real solutions to the equation, but this will give you a list to start with, anyway.
-Dan
Can't EVER do too much Physics!
-Dan
Hello, fdrhs!
The second one factors . . . if you hammer at it long enough.
$\displaystyle [2]\;\;2x^4 + x^3 - 24x^2 + 20x + 16 \:= \:0$
Factor "by grouping" . . .
$\displaystyle x^3(2x + 1) - 4(x^2 + 5x - 4) \:=\:0$
$\displaystyle x^3\underbrace{(2x + 1)} - 4\underbrace{(2x + 1)}(3x - 4) \:=\:0$
$\displaystyle (2x + 1)(x^3 - 4[3x-4]) \:=\:0\quad\Rightarrow\quad (2x + 1)(x^3 - 12x + 16)\:=\:0$
We find that $\displaystyle x = 2$ is a zero of cubic. .Hence, $\displaystyle (x - 2)$ is a factor.
And we have: .$\displaystyle (2x + 1)(x^3-12x + 16)\:=\:0$
. . . . . Then: .$\displaystyle (2x + 1)\overbrace{(x - 2)(x^2 + 2x - 8)}\:=\:0$
. . . . . And: .$\displaystyle (2x + 1)(x - 2)\overbrace{(x - 2)(x + 4)}\:=\:0$
Therefore, the four roots are: .$\displaystyle x\;=\;\text{-}\frac{1}{2},\:2,\:2,\:\text{-}4$
I do apologise, I was doing it at late hours of night while chatting on MSN with others about their weekends.
Thanks for picking it up guys, and yes, I do mean discriminants... and no it wasn't negative, therefore you CAN factorise it.
I certainly look like a little bit of a fool at the moment
Thanks guys.