Question

8. Find the polynomial function of degree 3 for the given conditions I.f(1) = 32 and x-2, x-3,x+1 are factors of the polynomial II.f(3) = -24 and 5,2,4 are zeros of the polynomial​

Answer 1

Part I

In factored form,
`f(x)=a(x-2)(x-3)(x+1)`. Substituting x=1,

`f(1)=32=a(1-2)(1-3)(1+1)\\\\\implies 32=4a\\\\a=8\\\\\therefore f(x)=8(x-2)(x-3)(x+1)`

Part II

In factored form,
`f(x)=a(x-5)(x-2)(x-4)`. Substituting x=3,

`f(3)=-24=a(3-5)(3-2)(3-4)\\\\\implies 24=2a\\\\a=12\\\\\therefore f(x)=12(x-5)(x-2)(x-4)`