Factoring question

• Jan 31st 2010, 01:05 PM
james121515
Factoring question
Hi,

What is the best method to go about factoring this \$\displaystyle x^6+6x^4+10x^2+8\$? Can this be done by grouping? I tried to do rational root tests but it appears as though it does not have any rational roots.

Thanks for your help,
James
• Jan 31st 2010, 01:13 PM
pickslides
Quote:

Originally Posted by james121515
Hi,

What is the best method to go about factoring this \$\displaystyle x^6+6x^4+10x^2+8\$? Can this be done by grouping? I tried to do rational root tests but it appears as though it does not have any rational roots.

Thanks for your help,
James

Might be helpful to say \$\displaystyle a = x^2\$ making the whole thing cubic instead.
• Jan 31st 2010, 01:16 PM
skeeter
Quote:

Originally Posted by james121515
Hi,

What is the best method to go about factoring this \$\displaystyle x^6+6x^4+10x^2+8\$? Can this be done by grouping? I tried to do rational root tests but it appears as though it does not have any rational roots.

Thanks for your help,
James

\$\displaystyle y = x^6+6x^4+10x^2+8 > 0\$ for all \$\displaystyle x\$ ... no real roots.
• Jan 31st 2010, 01:17 PM
Henryt999
Well
Thats an expression you have no = sign anywhere.
Also it doesn´t have any x intercepts so it doesn´t factor.
• Jan 31st 2010, 01:25 PM
james121515
It factors to \$\displaystyle (x^2+2)^3\$. I'm curious as to what method you use to do it.
• Jan 31st 2010, 01:33 PM
pickslides
Quote:

Originally Posted by james121515
It factors to \$\displaystyle (x^2+2)^3\$. I'm curious as to what method you use to do it.

\$\displaystyle x^6+6x^4+10x^2+8\$

make \$\displaystyle a = x^2\$ now

\$\displaystyle a^3+6a^2+10a+8\$

Then employee the factor theorem checking positive and negative factors of 8.

Spoiler:
\$\displaystyle a = -2\$