Question

Write a 4th degree polynomial with real coefficients and given roots; roots= 1, 2, 3+i

Answer 1

I got \[x^4-9x^3+24x^2-24x+8=0\] But it says that this answer is incorrect

Answer 2

The real roots work out, checking complex root now.

Answer 3

Ah doesn't work, well you need to multiple (x-1)(x-2)(x-3-i)(x-3+i) lets see and the last two multiply to a result of (x^2-3x+ix-3x+9-3i-ix+3i+1) or x^2-6x+10

Answer 4

so (x-1)(x-2)(x^2-6x+10)

Answer 5

ahh I see what I did thanks

Answer 6

If a polynomial has roots a, b, c, and d, then the simplest polynomial that has those roots is
(x - a)(x - b)(x - c)(x - d)

If a polynomial with real coefficients has complex roots, then the complex roots must come in pairs of complex conjugate roots.

The roots 1 and 2 are real roots, so you need to use the binomials (x - 1) and (x - 2).
Since the root 3 + i is complex, you must also have the roots 3 - i. This gives you the factors: (x - (3 + i)) and (x - (3 - i))

When you put it all together, youi must multiply:

(x - 1)(x - 2)(x - 3 - i)(x - 3 + i) to find the polynomial.