Thread: How do I find the imaginary number solutions to this problem?

So I have the following equation:

$\displaystyle
r^3 + 4r^2 - 24 = 0
$

I can find the real solution of 2 by using synthetic division and finding which of the multiples of 24 ends up to have no remainder.

But, the back of the book also wants an imaginary number solution and I am lost as to how to find one. Any help would be greatly appreciated.

$\displaystyle (r-2)(r^2+6r+12)=r^3+4r^2-24$.
Can you find the roots of $\displaystyle r^2+6r+12~?$

You can use the factor theorem to say that $\displaystyle r-2$ is a factor of $\displaystyle f(r)$

Hence $\displaystyle (r-2)(Ar^2+Br+C) = r^3+4r^2-24$ where A, B and C are constants to be found. Normally. I'd equate the coefficients but you can also do long division

edit: too slow

I wouldn't have even bothered with synthetic division. Factor theorem will work nicely.
You've established that r=2 is a root, therefore by the factor theorem, $\displaystyle (r-2)$ is a factor. So take this out (via factorization):
$\displaystyle
(r-2)(Ar^2+Br+C)=0$
Then just complete the square or use the formula. on the quadratic for the complex solution(s).

@Plato:

Yeah, plug into the the quadratic equation and you get:
$\displaystyle x = -6+/-sqrt(36 - 4(1)(12))/2(1)$
$\displaystyle x = -6+/-sqrt(36 - 48)/2$
$\displaystyle x = -6+/-sqrt(-12)/2$
$\displaystyle x = -6+/-2*sqrt(-3)/2$
$\displaystyle x = -6+/-2i*sqrt(3)/2$
$\displaystyle x = -3+/-i*sqrt(3)$

Thanks!