# Pre Calc

posted by on .

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.

• Pre Calc - ,

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) - ,

How do you know that it is supposed to have an a in front?

• Pre Calc - ,

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

• Zack Pre Calc - ,

I got -53/610 also

• Bob Dylan - Pre Calc - ,

" 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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