Question

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

Answer 1

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

Answer 2

so is this the formula i'm suppose to follow

Answer 3

yes play with it

Answer 4

alright

Answer 5

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

Answer 6

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

Answer 7

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 \[(x+1)(x+1)=0\\ f(x)=C(x-a)(x-b)(x-c)...\]

Answer 8

ohhh i think i got it thanks

Answer 9

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