# Polynomial Express

• Feb 19th 2008, 05:37 PM
nerdzor
Polynomial Express
Solve:

$x^3-5x^2+8x-4 = 0$

$x^3-5x^2+8x-4 > 0$
• Feb 19th 2008, 05:48 PM
Jhevon
Quote:

Originally Posted by nerdzor
Solve:

$x^3-5x^2+8x-4 = 0$

by the remainder theorem, we see that $x = 1$ is a root. thus by the factor theorem, $(x - 1)$ is a factor.

dividing our cubic by $x - 1$, we see that:

$x^2 - 5x^2 + 8x - 4 = (x - 1)(x^2 - 4x + 4) = (x - 1)(x - 2)^2 = 0$

thus, we have $x = 1$ or $x = 2$

Quote:

$x^3-5x^2+8x-4 > 0$
i leave this part to you. we found our roots to be $x = 1$ and $x = 2$, so we in fact can split up the real line into the intervals $(-\infty, 1)$, $(1,2)$, and $(2, \infty)$

on which of these intervals is the cubic positive?
• Feb 19th 2008, 05:52 PM
bobak
By inspection $(x-1)$ is a factor because $x = 1$ is a zero of the expression

then by use synthetic division

$x^3-5x^2+8x-4 = (x-1)(ax^2+bx+c)$

Compare coefficients of the highest order term to get a = 1 and lowest order to get c = 4 first because it is easy

$x^3-5x^2+8x-4 = (x-1)(x^2+bx+4)$ by comparing the coefficient of the term with x^2 or otherwise you should easily get b = -4

$x^3-5x^2+8x-4 = (x-1)(x^2-4x+4)$
$\Rightarrow x^3-5x^2+8x-4 = (x-1)(x-2)^2$

for the inequality you must sketch a graph.

You know be able to tell that for large positive values of x the function will be positive, and for large negative values of x the function will be negative.
You should also know the the sign of the function can only change at a root.

you should get a graph like this. the image below.
Remember the inequality is greater than zero not greater than or equal to, don't be lazy and just write x >1