1. solve for x, xER using algebraic methods. Use a table to show your results.
$\displaystyle x^3-2x^2-x+2 > 0$
Can someone please guide me on this one?
Hello,
Factor $\displaystyle x^3-2x^2-x+2$
you can see that 1 is a root of this polynomial. Thus (x-1) is a factor of the polynomial.
solve for a,b in $\displaystyle x^3-2x^2-x+2=(x-1)(x^2+ax+b)$
you'll get two roots, p and q :
$\displaystyle x^3-2x^2-x+2=(x-1)(x-p)(x-q)$
(p=-1 and q=2)
then make a table, like they said :
between - infinity and -1, (x-1)<0, (x+1)<0, (x-2)<0, so the polynomial is...
between -1 and 1, ...
between 1 and 2, ...
between 2 and infinity, ...
if you have no idea on how to do this, just say it here, but please try to think about it