What is the third–degree polynomial function such that f(0) = –24 and whose zeros are 1, 2, and 3?

Question

anonymous · Answer

$$f(x)=ax^3+bx^2+cx+d=(x-u)(x-v)(x-w)//f(0)=d$$ where u v w is the roots abcd is the coefficient

anonymous · Answer

so is this the formula i'm suppose to follow

anonymous · Answer

yes play with it

anonymous · Answer

alright

anonymous · Answer

well im stumped could you simplify it then ill solve it from there

anonymous · Answer

more hint then the zeros are 1,2,3 so 0=(x-1)(x-2)(x-3)

anonymous · Answer

remember, in the factored form a constant is lost, for example $$2x^2+4x+2=2(x^2+2x+1)=2(x+1)(x+1)$$ and $$(x+1)(x+1)=0\ f(x)=C(x-a)(x-b)(x-c)...$$

anonymous · Answer

ohhh i think i got it thanks

anonymous · Answer

and that constant is related very much to f(0)=-24, try to find that constant