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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)?
A. 48
B. 24
C. 0
D. 50
Read Solution (Total 5)
-
- Roots are 1,2,3,4
So we can write this
equation As p(x)=
a(x-1)(x-2)(x-3)(x-4).....eq1
Put x=0here we p(0)=48=24a
TheN a=2
Now put a in eq1
And get the answer...
P(5)=a.4.3.2.1=48 - 9 years agoHelpfull: Yes(8) No(0)
- 48 IS THE ANSWER
- 9 years agoHelpfull: Yes(2) No(1)
- can u explain how the ans is 48?
- 9 years agoHelpfull: Yes(2) No(0)
- 48 is the answer
- 9 years agoHelpfull: Yes(0) No(0)
- A. 48,
since 1,2,3,4 are the roots of p(x) so it can be equation p(x) i.e for p(x)=0 we get the value of x.so p(5)= 48 because only constant value "e" is evaluated and all other are zero. - 8 years agoHelpfull: Yes(0) No(0)
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