If P x ax 4 bx 3 cx 2 dx e has roots at x 1 2 3 4 and

p(0)=c(x-1)(x-2)(x-3)(x-4)
48=c(-1*-2*-3*-4)
c=2
p(5)=2(5-1)(5-2)(5-3)(5-4)
=2*4*3*2*1
=48
11 years agoHelpfull: Yes(2) No(0)
P(x)=ax^4+bx^3+cx^2+bx+e can be rewrite as P(x)=k(x-1)(x-2)(x-3)(x-4)
To find k value=>P(0)=48 so,P(0)=k(-1)(-2)(-3)(-4)=24k
24k=48
k=2
then P(5)=2(5-1)(5-2)(5-3)(5-4)=2*4*3*2*1=48
therefore P(5)=48
11 years agoHelpfull: Yes(1) No(0)
p(0) =48
so we have the four roots
(x-1) * (x-2) *(x-3) *(x-4) the four root
so p(0)= -1*-2*-3*-4 = 24 so we need 24
that p(x) =(x-1) * (x-2) *(x-3) *(x-4) +24
p(5)= (5-1) * (5-2) *(5-3) *(5-4)+24 = 48
9 years agoHelpfull: Yes(0) No(0)

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)