If p(x)=ax^4+bx^3+cx^2+dx+e has roots at x=1,2,3,4 and p(0)=48, what is p(5)	Answers:
A) 48
B) 24
C) 0
D) 50
E) 40
p(x)=ax^4+bx^3+cx^2+dx+e   it can be written as   p(x)=a*(x-1)*(x-2)*(x-3)*(x-4) ----(1)   put x=0 in(1), we get   p(0)=a*(-1)(-2)(-3)(-4)   =&gt; 48=a*24   =&gt; a=2   so, p(x)=2*(x-1)*(x-2)*(x-3)*(x-4)   =&gt; p(5)=2*(5-1)*(5-2)*(5-3)*(5-4)=2*4*3*2*1= 
<gadget id="calculator">2*4*3*2*1</gadget>
<output>48</output>
48   p(5)= 48   48NSWER:48
<result>A</result>
If p(x)=ax^4+bx^3+cx^2+dx+e has roots at x=1,2,3,4 and p(0)=48, what is p(5)