1.

Let P(x) be a cubic polynomial such that coefficient of `x^3` is 1 `, p(1) = 1, p(2) = 2, p(3) = 3 then the value of p(4) is :

Answer» Let `P(x) = x^3+ax^2+bx+c`
Then, `P(1) = 1+a+b+c = 1`
`=> a+b+c = 0->(1)`
`P(2) = 8+4a+2b+c = 2=> 4a+2b+c = -6->(2)`
`P(3) = 27+9a+3b+c = 3=>9a+3b+c = -24->(3)`
Subtracting (1) from (2)
`3a+b = -6->(4)`
Subtracting (2) from (3)
`5a+b = -18->(5)`
Subtracting (4) from (5)
`2a = -12 => a = -6`
`=> b = -6+18 = 12`
`c = -a-b = 6-12 = -6`
So, our polynomial becomes,
`P(x) = x^3-6x^2+12x-6 = 0`
`:.P(4) = 64-96+48-6 = 10`


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