Pre Calc
posted by Zach .
Write a quartic function that has roots that are 3+5i, 4 and 7 and
f(3)=53.
I have tried it a bunch of times and can't seem to get it.
(x3+5i)(x35i)(x+4)(x7)
I multiplied them together
x^49x^3+24x^2+66x952
I was a little unsure what f(3)=53 means.
I thought I should plug 3 in for x and it should equal 53. That doesn't work.
Can you please help me see where I went wrong?

Ahh, how would you know that the function wasn't
f(x) = a(x3+5i)(x35i)(x+4)(x7)
= a(x^49x^3+24x^2+66x952)
Wouldn't it still have those same roots if you set f(x) = 0 ?
I am going to trust that you expanded correctly.
Now let f(3) = 53
53 = a((3)^4  9(3)^3 + 24(3)^2 + 66(3)  952)
solve for a, then multiply your expanded expression by that value.
Take over... 
How do you know that it is supposed to have an a in front?

I get where a comes from. Is the answer 53/610? It doesn't seem right.

I got 53/610 also

" How do you know that it is supposed to have an a in front? "
Bob, I guess the answer was blowing in the wind.