Question

Find a third–degree polynomial function such that f(0) = –12 and whose zeros are 1, 2, and 3. Using complete sentences, explain how you found it.

Answer 1

Use factor-root theorem.
Factors are (x-1), (x-2), and (x-3)
You'll have to use the point (0,-12) to solve for the constant, a.

Answer 2

let take it simply
f(x)= ax^3+bx^2+cx+d
f(0)=d=-12 [as x=0]
f(1)=0[as it is factor]
so
a+b+c+d=0
f(2)=0
8a+4b+2c+d=0
f(3)=0
27a+9b+3c+d=0
you will get answer :)
@CliffSedge ur answer will not be true as you have to take a constant which after create problem:)

Answer 3

@Aperogalics , I don't know what you mean.
The third-degree function will have the form \(\large f(x)=a(x-1)(x-2)(x-3).\)
Simplifying, this becomes, \(\large f(x)=a(x^3-6x^2+11x-6).\)
\(f(0)=-12 \rightarrow -12=a(-6)\).

Can you point out my error, please?

Answer 4

You provided a general method, @Aperogalics which works given any four points to define the cubic, but when the zeroes are given, the work simplifies greatly.

Answer 5

@CliffSedge sry actually this thing is nt for you i m writing for somebody else wrote it here by mistake so don't mind it and i tell him general method so that he could attempt any type of question:)