Any polynomial equation having 1 and 3i as roots must have x- 1 and x- 3i as factors. In order that all coeffcients be rational (and, in particular, real) we must also have x+ 3i as a factor. Your polynomial is (x-1)(x-3i)(x+3i). Multply that out and see how the "i" terms disappear.

[quote] 2)-5 and 1 - iDoes it not bother you at all that that is NOT a third degree equation? That is NOT the correct answer.Same thing. The factors must be x-(-5)= x+ 5 and x- (1- i)= x-1+ i. In order that all coefficients be real there must also be a factor of x- (1+i)= x-1-i. (x+5)(x-1-i)(x-1+i). To see why we need to change the sign on i, first multiply (x-1- i)(x-1+i) as [(x-1)- i][(x-1)+ i]. Notice that that is a "sum and difference" multiplication.

Where did you get those formulas from? Where ever you saw them, it should have included what "r1", "r2", "a", and "b" mean. If you don't learn what all of the variables in a formulanote: i would appreciate if u use these two formulas and explain how to use them r1 + r2 = -b/a and r1 . r2 =c/a

i have this answered problem but i still dont get what the r1,r2,b,c and a representsmean, you haven't learned the formula.

Ithinkwhat you are referring to is a quadratic equation (and so it has nothing to do with this problem. If you have a polynomial equation of the form with roots and then and .

That's true because if and are roots of a polynomial equation, then and are factors of the polynomial. And since , to have "leading coefficient", a, we must multiply that by a: setting coefficients of the same powers equal, so and so r_1r_2= \frac{c}{a}[/tex].

But, as I say, this is true only for quadratic equations and the equations you have are cubic.

[qote]ex: Find a third degree polynomial equation with rational coefficients has the given numbers as roots 2+3i and 2-3i

answer) r1 + r2 = -b/a = -4/1

r1 . r2 = c/a = 13/1

the equation is x^2 - 4x + 13 =0

What I would do is say that since 2+ 3i and 2- 3i are roots, x- (2+ 3i) and x- (2-3i) must be factors. [quote](x-(2+3i))(x-(2-3i))= ((x-2)-3i)((x-2)+3i)= (x-2)^2- (3i)^2[/tex] is thequadraticequation having those numbers as roots. To get a cubic equation, with rational coefficients, with those roots, multiply that by (ax+ b) where a and b can be any rational numbers.

Again, your formulas are only true forquadraticequations, not cubic equations. By using the same idea, writing the equation as a product of factors you can say this:

If , , and are roots of the cubic polynomial equation , then , , and .

But thesimplestway to determine a polynomial equation, with real coefficients, with given roots is to write the factors as , , etc., including "complex conjugates" of any complex roots to "keep it real".