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...
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?
5 answers
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.
Bob, I guess the answer was blowing in the wind.
1 answer