Question 703877
AN PICTURE IS WORD A THOUSAND WORDS:
This is how my 11th grade math teacher taught me how to solve that kind of problem. (Teachers may have tried to teach me some math earlier than that, but I was not paying too much attention).
The sign of a product:
{{{drawing(500,300,-2,5,-1,5.2,
locate(-2,0.4,x^2(2x - 1)),locate(-0.5,1,number_line(350, -2, 3, 0, 1/2)),
locate(-2,2.3,(2x - 1)),locate(-0.5,3,number_line(350, -2, 3, -5,1/2)),
locate(-2,4.4,x^2),locate(-0.5,5,number_line(350, -2, 3, 0,-5)),
locate(-0.35,1,"-"),locate(0.15,1,"-"),locate(0.65,1,"-"),locate(1.15,1,"-"),
locate(1.4,0.95,0),
locate(1.65,1,"-"),locate(1.9,0.95,0),locate(2.2,1,"+"),locate(2.7,1,"+"),
locate(3.2,1,"+"),locate(3.7,1,"+"),locate(4.2,1,"+"),
locate(-0.35,3,"-"),locate(0.15,3,"-"),locate(0.65,3,"-"),locate(1.15,3,"-"),
locate(1.4,3,"-"),
locate(1.65,3,"-"),locate(1.9,2.95,0),locate(2.2,3,"+"),locate(2.7,3,"+"),
locate(3.2,3,"+"),locate(3.7,3,"+"),locate(4.2,3,"+"),locate(0.2,3,"-"),
locate(-0.35,5,"+"),locate(0.15,5,"+"),locate(0.65,5,"+"),locate(1.15,5,"+"),
locate(1.4,4.95,0),
locate(1.65,5,"+"),locate(1.9,5,"+"),locate(2.2,5,"+"),locate(2.7,5,"+"),
locate(3.2,5,"+"),locate(3.7,5,"+"),locate(4.2,5,"+")
)}}} You place pluses, minuses, and zeros on the stacked (lined up) number lines for each factor.
Then, you figure out the pluses, minuses, and zeros of the product and write them on  the bottom number line (the one for the product)
 
GRAPHS:
{{{x^2}}} {{{graph(300,300,-1.5,1.5,-0.5,0.5,x^2)}}} times {{{(2x-1)}}} {{{graph(300,300,-2,2,-5,5,2x-1)}}} equals {{{x^2(2x-1)}}} {{{graph(300,300,-1.5,1.5,-0.5,0.5,x^2(2x-1))}}}
 
AN EXPLANATION IN MANY WORDS:
{{{x^2}}} is zero for {{{x=0}}}
For all other values of {{{x}}}, {{{x^2>0}}} is positive,
and {{{x^2*(2x - 1) }}} will go along with {{{(2x-1)}}},
being negative for all {{{x<1/2}}},
except for {{{x=0}}}, where {{{(2x-1)<0}}},
but because {{{x^2=0}}}, {{{x^2(2x-1)=0}}}
At that point {{{x^2(2x-1)}}} refuses to go along with {{{(2x-1)}}}.