SOLUTION: Show that {{{p(x)=2x^3-4x^2+5x-6}}} has a zero between -1 and 2

Algebra ->  Graphs -> SOLUTION: Show that {{{p(x)=2x^3-4x^2+5x-6}}} has a zero between -1 and 2       Log On


   

Question 140328: Show that p%28x%29=2x%5E3-4x%5E2%2B5x-6 has a zero between -1 and 2



Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
p%28x%29=2x%5E3-4x%5E2%2B5x-6 Start with the given equation


p%28-1%29=2%28-1%29%5E3-4%28-1%29%5E2%2B5%28-1%29-6 Plug in x=-1


p%28-1%29=-17 Multiply and combine like terms


-------------

p%28x%29=2x%5E3-4x%5E2%2B5x-6 Go back to the given equation


p%282%29=2%282%29%5E3-4%282%29%5E2%2B5%282%29-6 Plug in x=2


p%282%29=4 Multiply and combine like terms



Since p%28-1%29=-17 (which is negative) and p%282%29=4 (which is positive), this means that the graph has crossed the x-axis. So there is a zero in between -1 and 2