If p(x)= x^4 +ax3+bx²+cx+d and P(1)= P(2)=p(3)=0 then find the value of p(4) +p(0).
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p (1) , p (2) , p (3) gain zero value its means 1,2,3 are root of equation
hence,
x^4+ax^3+bx^2+cx+d=(x-1)(x-2)(x-3)(x-k)
=x^4+(-6-k) x^3+(11+6k) x^2+(-6-11k) x+6k
compare LHS to RHS
(-6-k)=a
(11+6k)=b
(-6-11k)=c
6k=d
now
p (4)+p (0)=256+64a+16b+4c+d+d
now put value of a , b, c, and d from above .
=256-384+176-24
=432-408
=24
when we put a, b and c value
hence,
x^4+ax^3+bx^2+cx+d=(x-1)(x-2)(x-3)(x-k)
=x^4+(-6-k) x^3+(11+6k) x^2+(-6-11k) x+6k
compare LHS to RHS
(-6-k)=a
(11+6k)=b
(-6-11k)=c
6k=d
now
p (4)+p (0)=256+64a+16b+4c+d+d
now put value of a , b, c, and d from above .
=256-384+176-24
=432-408
=24
when we put a, b and c value
mysticd:
how do u get 24
when put value a, b, c and d which we find in term of k then I find 24
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