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

Algebra 2

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

Postby blu3fan » Tue Jan 30, 2007 12:29 pm

Find the roots of: z^3 + 6z^2 - 4z - 24 = 0
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Postby keentoknow » Tue Feb 27, 2007 5:57 am

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

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Re: Solve..

Postby mathpedia » Thu May 28, 2009 7:18 am

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


Consider the associated polynom: [tex]f(z) = z^3 + 6z^2 - 4z - 24[/tex]
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 [tex]( z^3 + 6z^2 -4z - 24 )[/tex] divisible with (z-2). Putting the fraction and making the division, you get:
[tex]\frac{(z^3 + 6z^2 -4z - 24)}{x-2} = z^2 + 8z + 12.[/tex] So,
[tex]z^3 + 6z^2 -4z - 24 = (x-2) * (z^2 + 8z + 12)[/tex]

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

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

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Postby Ivan » Tue Dec 29, 2009 7:45 am

[tex]z^3+6z^2-4z-24=0[/tex]

[tex](z^3+6z^2)-(4z+24)=0[/tex]

[tex]z^2(z+6)-4(z+6)=0[/tex]

[tex](z+6)(z^2-4)=0[/tex]

[tex]x=-4, x=\pm2[/tex]

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Postby Mrs. Dazz » Wed Sep 15, 2010 9:16 pm

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

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Postby Math Tutor » Thu Sep 16, 2010 1:05 am

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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