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.

note: 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 represents

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 formula

**mean**, you haven't learned the formula.

I

**think** what 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 $\displaystyle ax^2+ bx+ c= 0$ with roots $\displaystyle r_1$ and $\displaystyle r_2$ then $\displaystyle r_1+ r_2= -\frac{b}{a}$ and $\displaystyle r_1r_2= \frac{c}{a}$.

That's true because if $\displaystyle r_1$ and $\displaystyle r_2$ are roots of a polynomial equation, then $\displaystyle x- r_1$ and $\displaystyle x- r_2$ are factors of the polynomial. And since $\displaystyle (x- r_1)(x- r_2)= x^2- (r_1+ r_2)x+ r_1r_2$, to have "leading coefficient", a, we must multiply that by a: $\displaystyle ax^2+ bx+ c= a(x- r_1)(x- r_2)= ax^2- a(r_1+ r_2)+ ar_1r_2$ setting coefficients of the same powers equal, $\displaystyle b= -a(r_1+ r_2)$ so $\displaystyle r_1+ r_2= -\frac{b}{a}$ and $\displaystyle c= ar_1r_2$ 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