Writing a minimum degree polynomial given roots

Question

Write a polynomial function of least degree with integral coefficients the zeros of which include 4 and 1-i.

I understand that the opposite of 1-i is implied do to it being imaginary, but do not understand how to write it. I assume it would be written as (x-4)(x+1-i)(x-1+i), but when I do this I get a polynomial that includes i, something that is not integral. The correct factors are (x-4)(x-1+i)(x-1-i) but I am having trouble figuring out why the sign of i is the flipped but the sign of 1 is not. Is it because only i is insinuated as appearing twice? Any help on this would be greatly appreciated.

Answer 1

Neol,

it is not the opposite of 1-i, but the CONJUGATE, which means the sign only changes for the imaginary component. So 1-i and 1+i are the two complex roots, while 4 is the only real root.

(x-4) [x-(1-i)][x-(1+i)]

"FOIL" the two complex factors first: x*x-x(1+i)-(1-i)x+(1-i)(1+i)

=x²-x-xi-x+xi+1+i-i-i²

Notice: all of the "i" terms cancel and the i² = -1, so it is no longer imaginary.

=x²-2x+2 NO i's, all integer coefficients.

Now, still need to multiply that by the (x-4)

(x-4)(x²-2x+2) = your polynomial

f(x)= x³-2x²+2x-4x²+8x-8

f(x)=x³-6x²+10x-8

Answer 2

The direct answer to your question is that complex roots of polynomial equations with real coefficients occur in conjugate pairs, i.e. if a + bi is a root then so is a - bi.

The polynomial from this question is (x - 4)(x - 1 - i)(x - 1 + i)

That is (x - 4)(x² - 2x + 2)

I'll leave that last multiplication to you.