Question 953541
The function {{{p}}} is a fourth-degree polynomial with x-intercepts {{{0.5}}}, {{{5}}}, and {{{10}}} and y-intercept {{{-1}}}. 
If {{{p(x)}}} is positive only on the interval ({{{5}}}, {{{10}}}), find {{{p(x)}}}. 

A fourth-degree polynomial must have {{{four}}} roots. {{{Three}}} are given. Presumably {{{one}}} of those three has multiplicity of {{{2}}}, as opposed to there being an entirely separate root that the problem isn't telling about.

{{{5}}} and {{{10}}} can't either have a multiplicity of {{{2}}} without violating the condition about "positive only on the interval ({{{5}}}, {{{10}}})", so that leaves {{{0.5}}} or {{{1/2}}} to be the multiple root.

Use the factor theorem to convert the four roots into factors:

 {{{p(x)=(x -1/2) (x-1/2) (x -5) (x -10)}}}

Multiply these factors together:

 {{{p(x)=x^4-16x^3+(261x^2)/4-(215x)/4+25/2}}}

The y-intercept of this function is {{{25/2}}}, so to achieve the desired y-intercept, multiply all the coefficients by {{{(-1 / (25/2)) = -2/25}}}:

{{{ p(x)= -(2/25)x^4-(-2/25)16x^3+(-2/25)(261x^2)/4-(-2/25)(215x)/4+(-2/25)25/2}}}

 {{{p(x)=-(2/25)x^4+(32/25)x^3+(-261/50)(x^2)+(215/50)x-1}}}

 {{{p(x)=-(2x^4)/25+(32x^3)/25-(261x^2)/50+(43x)/10-1}}}


{{{ graph( 600, 600, -15, 25, -15, 25,-(2x^4)/25+(32x^3)/25-(261x^2)/50+(43x)/10-1) }}} 


as you can see on a graph, 
x-intercepts  are {{{0.5}}}, {{{5}}}, and {{{10}}} 
y-intercept is {{{-1}}}, 
and 
{{{p(x)}}} is positive only on the interval ({{{5}}}, {{{10}}})