Finding Polynomial Functions with Complex Conjugates

[zoggero]

I'm getting stuck on this problem:
Find an equation of f(x), the polynomial of smallest degree, with coefficients, such that f(x) breaks through the x-axis at -5, bounces off the x-axis at 4, has the complex roots of 1-i and 1+5i, and passes through the point (0,74).
f(x)= ?

I understand so far that I have the roots (x+5)(x-4)^2(x-1-i)(x-1+5i)then according to the complex conjugate theorem I would also have (x-1+i) and (x-1-5i) I am unsure where to go from there. My assignment requires me to foil out the i's so I am not dealing with imaginary numbers and evaluate x at 0 when y is 74, but I am unclear on the steps I need to take. Any help would be appreciated.

 

[Deleted member 4993]

I'm getting stuck on this problem:
Find an equation of f(x), the polynomial of smallest degree, with coefficients, such that f(x) breaks through the x-axis at -5, bounces off the x-axis at 4, has the complex roots of 1-i and 1+5i, and passes through the point (0,74).
f(x)= ?

I understand so far that I have the roots (x+5)(x-4)^2(x-1-i)(x-1+5i)<<< Correct

then according to the complex conjugate theorem I would also have (x-1+i) and (x-1-5i) <<< correct - so include those

Then insert a multiplicative constant (A*(x+5)(x-4)^2.....) to account for the y-intercept

 

[stapel]

I am unclear on the steps I need to take. Any help would be appreciated.

For an explanation of the "A" mentioned in the previous reply, try here.

Short version: There are infinitely-many polynomials with a given set of zeroes, since A(p(x)) = 0 solves in exactly the same way as p(x) = 0. To find "the" polynomial with that set of zeroes, they give you a point (x, y) that you can plug into A(p(x)) = y, so you can solve for the value of A.

In your case, you'll be working with:

. . . . .\(\displaystyle A(x\, -\, 4)^2 (x\, +\, 5)(x^2\, -\, 2x\, +\, 2)(x^2\, -\, 2x\, +\, 26)\, =\, y\)

(Note: You might find it helpful to multiply the complex-values factors first, and do the multiplication vertically. It often seems easier to keep track of your work that way.)

Plugging in the given x- and y-values from the point they gave you, we get:

. . . . .\(\displaystyle A(-4)^2 (5)(2)(26)\, =\, 74\)

Solve to find the value of A.