Find a third-degree polynomial equation with rational coefficients that has the given numbers as roots. 1 and 3i
1 answer:
You might be interested in
It's going to be kind of crazy, but you need to use Pythagorean's Theorem for this. That will look like this: 
. FOIL out the left side to get 
. FOIL out the first of the 2 expressions on the right to get 
, and the second of the 2 to get 4x. Our equation now looks like this: 
. Combine like terms to get an equation that still has square roots in it that we have to deal with: 
and 
. We will square both sides to get rid of the square root sign. 
. This is a polynomial now that can be factored to solve for x. Bring the 2x over by subtraction and set the polynomial equal to 0. 
. Factor out an x, leaving us with x(x-2)=0. That means that x = 0 or x - 2 = 0 and x = 2. Of course if we are solving for the length of a side we know it can't have a side length of 0, so it must have a side length that is a multiple of 2. x = 2