I have made a start, I am a little stuck in the middle:

w = 2 – 3i is a solution of the equation
$$ 3w^3-14 w^2+Aw– 26 = 0$$
where A is real. Find the value of A and the other two solutions of the equation

Question

I have made a start, I am a little stuck in the middle:

w = 2 – 3i is a solution of the equation
$$ 3w^3-14 w^2+Aw– 26 = 0$$
where A is real. Find the value of A and the other two solutions of the equation

marigirl · Answer

I know the solution has to be a conjugate pair, so  I did

(w-(2+3i) )(w-(-2-3i) 

and got

$$w^2-4w+13 $$

a bit stuck from here............plz help :)

marigirl · Answer

OMGGGG I SOLVED IT!!!!!!!!!!!

marigirl · Answer

:D :D Thank guys for dropping by and checking out the question!

marigirl · Answer

If anyone is interested in knowing..

I mirrored w^2-4w+13 to look like my original cubic by multiplying it with (kw+c)

then matched up the coefficients to the values i have and then found the values i didnt have (i.e A)

marigirl · Answer

@LynFran  thanks for the medal :D :D :D

irishboy123 · Answer

you could also have done long division and set A so that the remainder is zero