Question 167067
Distributive property
Let's look at Z(C+D)=ZC+ZD.
What if Z=A+B?
Then 
Z(C+D)=(A+B)(C+D)=ZC+ZD=(A+B)C+(A+B)D=AC+BC+AD+BD
So then,
{{{(x+5)(x+2)=x*x+5x+2x+10=x^2+7x+10}}}
and
{{{(x-1)(x-3)=x*x+(-3)x+(-1)x+3=x^2-4x+3}}}
This is the FOIL method for expanding quadratic equations from factors.
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Now the problem becomes,
{{{p(x)=(1/2)(x+5)(x+2)(x-1)(x-3) =(1/2)(x^2+7x+10)(x^2-4x+3)}}}
Let's look at the product of the two quadratics,
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We again have to use the distributive property and make sure we account for each term. 
There are nine terms in all.
{{{(x^2+7x+10)(x^2-4x+3)=
x^2*(x^2-4x+3)+
7x*(x^2-4x+3)+
10*(x^2-4x+3)}}}
{{{(x^2+7x+10)(x^2-4x+3)=
x^2*(x^2)+x^2(-4x)+x^2(3)+
7x*(x^2)+7x(-4x)+7x(3)+
10*(x^2)+10(-4x)+10(3)}}}
{{{(x^2+7x+10)(x^2-4x+10)=
x^4-4x^3+3x^2+
7x^3-28x^2+21x+
10x^2-40x+30}}}
{{{(x^2+7x+10)(x^2-4x+10)=
x^4+(-4+7)x^3+(3-28+10)x^2+(21-40)x+30}}}
{{{(x^2+7x+10)(x^2-4x+10)=
x^4+3x^3-15x^2-19x+30}}}
Now we'll put that back in our original function.
{{{p(x)=(1/2)(x^4+3x^3-15x^2-19x+30)}}}
{{{p(x)=(1/2)x^4+(3/2)x^3-(15/2)x^2-(19/2)x+15)}}}

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Check, as an exercise, that we didn't make any mistakes and that x= -5,-2,1, and 3 are in fact zeros of this polynomial.
{{{p(0)=15}}}
So that checks out.