Question

Given that f(x) is a cubic function with zeros at -1 and -3 i - 4 , find an equation for f(x) given that f(7) = 7.

Answer 1

Since once of our roots is complex, we can automatically get our 3rd missing root by knowing that complex roots come in conjugate pairs. The other part of the conjugate pair is going to be the same root, but with the sign of the imaginary portion changed, meaning out 3rd root is 3i-4. So since these 3 are roots, we can use them to write our polynomial in terms of factors.
\[(x-(-1))(x-(3i-4))(x-(-3i-4)) = (x+1)(x-3i+4)(x+3i+4)\]
Now when constructing an appropriate polynomial, we want to put some sort of unknown constant multiple out in front, because it is very unlikely that the function we're looking for is only factors. So that means I want:
\[f(x) =A(x+1)(x-3i+4)(x+3i+4)\]
Now we can get A by using the condition we have, f(7) = 7. This is basically saying that when x = 7, y must also = 7. I can plug those two values in now and see what A is.
\[7 = A(7+1)(7-3i+4)(7+3i+4)\]
\[7=A(8)(11-3i)(11+3i)\]
\[7 = A(8)(121+9) \]
\[7 = A(1040) \implies A = \frac{ 7 }{ 1040 }\]So that means the polynomial we want is:
\[f(x) = \frac{ 7 }{ 1040 }(x+1)(x-3i+4)(x+3i+4)\]
Now we just foil this out to get our full polynomial.
\[f(x) = \frac{ 7 }{ 1040 }(x^{3} + 9x^{2} + 33x + 25)\]
If you want to see how that was foiled out feel free to ask, but that would be your final answer.

Answer 2

ty i got it :)