Question

Find a fourth degree polynomial with real coefficients that has zeros of -3, 2, i that passes through (-2, 100).

Answer 1

`f(x) = -5x^(4) -5x^(3) +25x^(2) -5x+30`

Explanation:

A fourth degree polynomial is of the form

f(x) = K(x-a)(x-b)(x-c)(x-d)

Where K is a constant and a,b,c, and d are roots of the equation.

We have three roots already: -3, 2, i

Because we are told that the coefficients are real and we have an imaginary zero, i, we need to obtain its conjugate which is = -i

So, the four roots are: a = -3, b = 2, c = i, d = -i

f(x) = K(x-(-3))(x-2)(x-i)(x-(-i))

f(x) = K(x+3)(x-2)(x-i)(x+i)

`f(x)=K(x^(2) -2x+3x-6)(x^(2) -ix+ix-i^(2))`

But
`i^(2) = -1`

`f(x) = K(x^(2)+x-6)(x^(2) +1)\\\\f(x) = K(x^(4)+x^(3) -6x^(2) + x^(2) +x-6)\\\\f(x) = K(x^(4) +x^(3) - 5x^(2) +x-6)`

The polynomial passes through (-2, 100). That is, at x = -2, f(x) = 100

`100 = K[(-2)^(4)+(-2)^(3) -5(-2)^(2)+(-2)-6]\\100 = K(16-8-20-2-6)\\100 = -20K\\\\K = -5`

`f(x) = -5(x^(4) +x^(3) - 5x^(2) +x-6)\\f(x) = -5x^(4) -5x^(3) +25x^(2) -5x+30`