Question 580756
If {{{(x-4)}}}, or {{{(x-4)^3}}}, or  {{{(x-4)^3}}}, or ... (well, you get the idea) is a factor of a polynomial, the polynomial will be zero AND change sign at x=4. Then the polynomial "crosses the x-axis at ... 4". To make it the simplest, we should include
{{{(x+1)(x-4)}}} as a factor in the polynomial, so that it crosses the x-axis at -1 and 4."
On the other hand if we include {{{x^2}}} and {{{(x-2)^2}}} as factors in the polynomial, the polynomial will be zero, but will not change sign at x=0 and x=2, so we would say that it touches (but does not cross) the x-axis.
So {{{f(x)=x^2(x-2)^2(x+1)(x-4)}}} would be a polynomial that "crosses the x-axis at -1 and 4," and "touches the x-axis at 0 and 2."
How does it behave between 0 and 2? If it is positive, then it "is above the x-axis between 0 and 2."
But {{{f(1)=1^2(1-2)^2(1+1)(1-4)=1*(-1)^2*2*(-3)=1*1*2*(-3)=-6<0}}}, so we need a (-1) as an extra factor.
{{{P(x)=-x^2(x-2)^2(x+1)(x-4)}}} is a polynomial that complies with the problem's requirements.
{{{graph(300,300,-1.5,4.3,-75,75,-x^2(x-2)^2(x+1)(x-4))}}}