# Thread: Find a third degree polynomial equation with the given roots?

1. ## Find a third degree polynomial equation with the given roots?

hi there,
this is algebra two honors question

Find a third degree polynomial equation with rational coefficients has the given numbers as roots
1) 1 and 3i 2)-5 and 1 - i

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

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

2. Originally Posted by mustafa
hi there,
this is algebra two honors question

Find a third degree polynomial equation with rational coefficients has the given numbers as roots
1) 1 and 3i
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 - i
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 $ax^2+ bx+ c= 0$ with roots $r_1$ and $r_2$ then $r_1+ r_2= -\frac{b}{a}$ and $r_1r_2= \frac{c}{a}$.

That's true because if $r_1$ and $r_2$ are roots of a polynomial equation, then $x- r_1$ and $x- r_2$ are factors of the polynomial. And since $(x- r_1)(x- r_2)= x^2- (r_1+ r_2)x+ r_1r_2$, to have "leading coefficient", a, we must multiply that by a: $ax^2+ bx+ c= a(x- r_1)(x- r_2)= ax^2- a(r_1+ r_2)+ ar_1r_2$ setting coefficients of the same powers equal, $b= -a(r_1+ r_2)$ so $r_1+ r_2= -\frac{b}{a}$ and $c= ar_1r_2$ 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
Does it not bother you at all that that is NOT a third degree equation? That is NOT the correct answer.

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] $x^2- 4x+ 4-(-9)= x^2- 4x+ 13= 0$ is the quadratic equation 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 for quadratic equations, not cubic equations. By using the same idea, writing the equation as a product of factors you can say this:
If $r_1$, $r_2$, and $r_3$ are roots of the cubic polynomial equation $ax^3+ bx^2+ cx+ d= 0$, then $r_1+ r_2+ r_3= -\frac{b}{a}$, $r_1r_2+ r_1r_3+ r_2r_3= \frac{c}{a}$, and $r_1r_2r_3= -\frac{d}{a}$.

But the simplest way to determine a polynomial equation, with real coefficients, with given roots is to write the factors as $x- r_1$, $x- r_2$, etc., including "complex conjugates" of any complex roots to "keep it real".