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March 29, 2017

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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 - ,

    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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