Polynomial function

Question

Write a polynomial function with real coefficients that satisfies the given condition: degree 3; zeros are 2 and 1-4i

Answer 1

Two of 3 possible roots were given for this polynomial of degree 3. Since the coefficients must be real, the other root must be the conjugate of the given complex root. We can now write a function in factored form:

(x - 2)(x - (1 - 4i))(x - (1 + 4i)) = 0

(x - 2)(x² - (1 + 4i + 1 - 4i)x + (1 - 4i)(1 + 4i)) = 0

(x - 2)(x² - 2x + 17) = 0

Or reverse engineer the quadratic equation:

For y = ax² + bx + c = 0, the roots will be [-b +/- sqrt(b² - 4ac)]/2a

Let a = 1. We seek b and c for roots = 1 +/- 4i

-b/2 = 1, so b = -2

[sqrt(4 - 4c)]/2 = 4i

sqrt(1 - c) = 4i

1 - c = -16

c = 17

This agrees with the first method.

So now the polynomial can be written in standard form by multiplying (x - 2)(x² - 2x + 17)

x³ - 4x² + 21x - 34

Any other polynomial that's a multiple of this (with real factors) could also work.

Answer 2

Interesting problem. Since the degree is 3, you would expect 3 roots (whether singly, or as multiple). But you only have 2. So, one of those roots must be a double. But it can't be the 2 (roots of (x-2), (x-2)), since the remaining root would be (x-(1-4i)), and there would be complex coefficients. But also, it can't be the (1-4i), since that would directly give a complex term: (x-(1-4i))*(x-(1-4i)) = (x^2-2x+8ix+15+8i) [I think!]. Now, if that "2nd" root was 1+/- 4i, then we can do business!

Maybe the problem was just asking you to reason it through like that?

-- Cheers, -- Mr. d.