Question 983489
{{{f(x)=(x+4)(x+1)(x-5)(x-a)}}}
This satisfies the three zeros. 
(-4,0)
(-1,0)
(5,0)
Now solve for {{{a}}}.
{{{f(0)=(0+4)(0+1)(0-5)(0-a)=5}}}
{{{4(1)(-5)(-a)=5}}}
{{{4a=1}}}
{{{a=1/4}}}
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{{{f(x)=(x+4)(x+1)(x-5)(x-1/4)}}}
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*[illustration fd11.JPG].
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You can also multiply by any constant to get another function.

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{{{g(x)=5(x+4)(x+1)(x-5)(x-1/4)}}}
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