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

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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

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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 :)
OMGGGG I SOLVED IT!!!!!!!!!!!
:D :D Thank guys for dropping by and checking out the question!

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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)
@LynFran thanks for the medal :D :D :D
you could also have done long division and set A so that the remainder is zero

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