Question

Find a polynomial p of degree 3 such that -1, 3 and 6 are zeros of p and p(0)=1

Answer 1

thought I had this one but my P(0) is 18. can you confirm that hthe problem is P(0) = 1

Answer 2

I am positive that p(0)=1

Answer 3

:(

Answer 4

y = 1/18(x+1)(x-3)(x-6)

Answer 5

y = (1/18) x^3 - (4/9) x^2 + (1/2) x + 1

Answer 6

(x +1) (x -3) (x -6)
(x +1) (x^2 - 9x + 18)
x^3 - 9x^2 + 18x + (x^2 - 9x + 18) + 1
x^3 - 8x^2 + 9x + 19

Answer 7

Cant we do it, by taking a polynomial of form, \[ax ^{3}+bx ^{2}+cx+d\] and equating it to the various values given? I am getting rather odd values for a,b,c,d. But atleast the process is right

Answer 8

p(x) = a(x+1)(x-3)(x-6)
p(0) = a(1)(-3)(-6) = 1
18a = 1
a = 1/18

p(x) = 1/18(x+1)(x-3)(x-6) = (1/18)x^3 - (4/9)x^2 +(1/2)x +1