Question 1168615
What is the polynomial function with the lowest degree whose some of its 

given:

zeros are {{{x[1]=-1}}},{{{x[2]=2}}},{{{x[3]=-6}}}-> three zeros tells you that you have {{{3}}}rd degree function, and intersections are at points ({{{-1}}},{{{0}}}),({{{2}}},{{{0}}}), and ({{{-6}}},{{{0}}})

use zero product rule to find it

{{{f(x)=(x-x[1])(x-x[2])(x-x[3])}}}...........substitute zeros

{{{f(x)=(x-(-1))(x-2)(x-(-6))}}}

{{{f(x)=(x+1)(x-2)(x+6)}}}....multiply

{{{f(x)=(x^2 - x - 2)(x+6)}}}

{{{f(x)=x^3 + 5x^2-8x-12}}}->your function


check the graph

{{{drawing( 600, 600, -15, 15, -15, 15, 
circle(-1,0,.12),circle(2,0,.12),circle(-6,0,.12),
locate(-1,1,p(-1,0)),locate(-6,1,p(-6,0)),locate(2,1,p(2,0)),
 graph( 600, 600, -15, 15, -15, 15, x^3 + 5x^2-8x-12)) }}}