Posted by **Zach** on Sunday, November 15, 2009 at 9:30pm.

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.

(x-3+5i)(x-3-5i)(x+4)(x-7)

I multiplied them together

x^4-9x^3+24x^2+66x-952

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?

- Pre Calc -
**Reiny**, Sunday, November 15, 2009 at 10:12pm
Ahh, how would you know that the function wasn't

f(x) = a(x-3+5i)(x-3-5i)(x+4)(x-7)

= a(x^4-9x^3+24x^2+66x-952)

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...

- Pre Calc (please explain) -
**Bob Dylan**, Sunday, November 15, 2009 at 10:23pm
How do you know that it is supposed to have an a in front?

- Pre Calc -
**Zach**, Sunday, November 15, 2009 at 10:47pm
I get where a comes from. Is the answer 53/-610? It doesn't seem right.

- Zack Pre Calc -
**Reiny**, Sunday, November 15, 2009 at 10:56pm
I got -53/610 also

- Bob Dylan - Pre Calc -
**Reiny**, Sunday, November 15, 2009 at 10:59pm
" How do you know that it is supposed to have an a in front? "

Bob, I guess the answer was blowing in the wind.

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