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Mathematics 68 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

so is this the formula i'm suppose to follow

OpenStudy (anonymous):

yes play with it

OpenStudy (anonymous):

alright

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

ohhh i think i got it thanks

OpenStudy (anonymous):

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

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