# Solve.., Find the roots of: z^3 + 6z^2 - 4z - 24 = 0

Algebra

### Solve.., Find the roots of: z^3 + 6z^2 - 4z - 24 = 0

Find the roots of: z^3 + 6z^2 - 4z - 24 = 0

blu3fan

Posts: 5
Joined: Sun Jan 21, 2007 4:11 am

Rewrite it as z^2(z+6)-4(z+6)=0
keentoknow

Posts: 7
Joined: Sun Feb 25, 2007 1:21 pm

### Re: Solve..

blu3fan wrote:Find the roots of: z^3 + 6z^2 - 4z - 24 = 0

Consider the associated polynom: $f(z) = z^3 + 6z^2 - 4z - 24$
It' s easy to see that f(2) = 2^3 + 6*2 ^2 - 4*2 - 24 = 8 + 36 - 8 - 24 = 0.
From Bezout theorem you have that if a is a root of a polynom anX^n + ... + a1x + a0, with a0, a1, ..., an real coefficients, then that polynom is divided by (X-a).

So $( z^3 + 6z^2 -4z - 24 )$ divisible with (z-2). Putting the fraction and making the division, you get:
$\frac{(z^3 + 6z^2 -4z - 24)}{x-2} = z^2 + 8z + 12.$ So,
$z^3 + 6z^2 -4z - 24 = (x-2) * (z^2 + 8z + 12)$

We can find the other 2 solutions of your equation, by finding the roots of $z^2 + 8z + 12$:
$z2,3 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2z}$, where we have $az^2 + bz + c = 0$, where a,b, c in R, a not Null.

On our example, a = 1, b = 8, c = 12.=>
$z2,3= \frac{-8 \pm \sqrt{8^2 - 4*1*12}}{2*1} =$
$\frac{-8 \pm \sqrt{64 - 48}}{2} = \frac{-8 \pm \sqrt{16}}{2} =$
$\frac{-8 \pm \4}{2} => z1 = \frac{-8+4}{2} = \frac{-4}{2} => z2 =-2$
$z3 = \frac{-8 - 4}{2} = \frac{-12}{2} => z3 = -6.$
mathpedia

Posts: 2
Joined: Thu May 28, 2009 6:04 am

$z^3+6z^2-4z-24=0$

$(z^3+6z^2)-(4z+24)=0$

$z^2(z+6)-4(z+6)=0$

$(z+6)(z^2-4)=0$

$x=-4, x=\pm2$
Ivan

Posts: 1
Joined: Tue Dec 29, 2009 7:40 am

I have a problem, can someone help? Please show your work so I can understand it.

Subtract. Simplify by collecting like radical terms.

2√75 - 4√3
Mrs. Dazz

Posts: 1
Joined: Wed Sep 15, 2010 9:08 pm

Here is the solution:
2√75 - 4√3 = 2√(3.25) - 4√3 = 2.√3.√25 - 4√3 = 2.5.√3 - 4√3 = √3(10 - 4) = 6√3
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