A polynomial in `x` of degree 3 vanishes when `x=1` and `x=-2` , ad has the values 4 and 28 when `x=-1` and `x=2` , respectively. Then find the value of polynomial when `x=0` .

Explanation

From the given information
`f(x) = (x-1)(x+2)(ax+b)`
Now `" "f(-1) = 4 and f(2) = 28 . Thus ,`
`(-1 -1) (-1 + 2) (-a + b) = 4`
and `(2-1)(2+2)(2a+b)=28`
`rArr a - b = 2 and 2a + b = 7`
`ax^(2) + bx + c - p = p`
`or " " a(x- alpha )(x-beta)= p` " [Using (1)]
Solving, a= 3 and b = 1
`rArr f(x) = (x-1)(x+2)(3x + 1)`
`rArr f(0) = -2`