Question 959245
given:
a fifth-degree polynomial with
the x-intercepts {{{-5}}}, {{{2}}}, and {{{10}}}, 
y-intercept {{{500}}} and
{{{f(x) >= 0}}} for {{{x<= 10 }}}

By hypothesis, {{{f(x)}}} has factors {{{(x -(-5))}}}, {{{(x - 2)}}}, and{{{ (x-10)}}} by the Factor Theorem.

==> {{{f(x) = (x + 5)(x -2)(x-10) *g(x)}}} for some 2nd degree polynomial {{{g(x)}}}.

To determine {{{g(x)}}}:

We need {{{f(x)>= 0}}} for {{{x <= 10}}}.

==> {{{ (x + 5)(x -2)(x-10) *g(x)>= 0}}}  for {{{x <= 10}}}

Note that each factor is negative for some {{{x <= 10}}}.
There a few ways to deal with this; here is one way.

Note that both {{{x + 5}}} and {{{x - 2}}} change signs when {{{x <= 10}}}, while {{{x -10}}} does not.

Since we need {{{g(x)}}} to be quadratic, we can take {{{g(x) = A(x + 5)(x -2)}}} for some constant {{{A}}}.


Now, we have {{{f(x) = A(x + 5)^2 (x- 2)^2 (x -10)}}} for some {{{A}}}.

Note: that we have not changed the x-intercepts by repeating some of the factors 

Since {{{(x + 5)^2 (x -2)^2}}} is never negative, needing {{{f (x) >= 0}}} for {{{x <=10}}} reduces to needing {{{A(x -10) >=0}}} for {{{x<=10}}}.

*This is guaranteed if {{{A < 0}}}.

Finally, we use {{{f(0) = 500}}} to determine {{{A}}}:
{{{A * 5^2 (-2)^2 (-10) = 500}}}
{{{A * 25 *4 (-10) = 500}}}
{{{A * (-1000) = 500}}}
{{{A  = 500/-1000}}}
==>{{{ A = -1/2}}}, which is indeed negative.

Hence, we can take 

{{{f(x) = (-1/2)(x + 5)^2 (x - 2)^2 (x - 10)}}}

{{{f(x) =(-1/2)(x^5-4 x^4-71 x^3+50 x^2+700 x-1000)}}}

{{{f(x) =-x^5/2+2 x^4+(71x^3)/2-25x^2-350x+500}}}