Factorizing Problem

**BobBali** · May 16th 2012, 12:59 AM

Hello All,

Need a little help finding the roots for:

$m^6 - 8$
Let $A= x^2$
$(a)^3 - 8= 0$

$(a)^3 = 8 Therefore, a^3 = 8 so, a^2 = 8 hence, (x^2 - 2) is one root$

$(x^2 - 2) (X^4 + bx + 4) = 0$

After this how do i find the 'b' coefficient...?

**Reckoner** · May 16th 2012, 04:03 PM

Originally Posted by BobBali

$m^6 - 8$

This is a difference of cubes:

$m^6-8 = \left(m^2\right)^3 - 2^3 = \left(m^2-2\right)\left(m^4+2m^2+4\right)$

Set each factor equal to zero. The second factor is actually quadratic in form, so we can use the quadratic formula to find its roots.

$m^6-8=\left(m^2-2\right)\left[\left(m^2\right)^2 + 2\left(m^2\right)+4\right]$

So

$m^2-2 = 0\Rightarrow m=\pm\sqrt2$

and

$\left(m^2\right)^2 + 2\left(m^2\right)+4=0\Rightarrow m^2 = \frac{-2\pm\sqrt{-12}}2$

$\Rightarrow m^2 = -1\pm i\sqrt3$

$\Rightarrow m = \pm\sqrt{-1+i\sqrt3}\mathrm{\quad or\quad}m=\pm\sqrt{-1-i\sqrt3}$

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