Solve:

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

$\displaystyle x^3-5x^2+8x-4 > 0$

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- Feb 19th 2008, 04:37 PMnerdzorPolynomial Express
Solve:

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

$\displaystyle x^3-5x^2+8x-4 > 0$ - Feb 19th 2008, 04:48 PMJhevon
by the remainder theorem, we see that $\displaystyle x = 1$ is a root. thus by the factor theorem, $\displaystyle (x - 1)$ is a factor.

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

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

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

Quote:

$\displaystyle x^3-5x^2+8x-4 > 0$

on which of these intervals is the cubic positive? - Feb 19th 2008, 04:52 PMbobak
By inspection $\displaystyle (x-1)$ is a factor because $\displaystyle x = 1$ is a zero of the expression

then by use synthetic division

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

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

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

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

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

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