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I'm looking at some textbook question and there is a part where it writes this: $[ 16\cos^4\theta-10\cos^2\theta+1]$ as this: $ \left( 2\cos^2 \theta-1 \right) \left( 8\cos^2 \theta-1 \right)$

So that it can be solved out for $\cos\theta$.

I know what the correct answer is but I'm trying to find out how to factorize similar cases. Did they use any trigonometric identity, or formula, that can help me?

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    I didn't write any of this myself, I just exported the latex from maple that's why2012-04-29

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Note the powers of $\cos$ in the first expression are $4$ and $2$. This would suggest that you have a quadratic expression in the variable $\cos^2\theta$. And indeed you do: set $z=\cos^2\theta$ if you like and write $16\cos^4\theta-10\cos^2\theta +1$ as $ 16(\cos^2\theta)^2 -10(\cos^2\theta)+1=16z^2-10z+1. $ Now you can factor $ 16z^2 -10z+1=(2z-1)(8z-1). $ Writing the above expression in terms of $\cos^2\theta$ gives you your second expression.

In general, you can recognize $az^4+bz^2+c$ as a quadratic expression in the variable $z^2$, or $az^6+bz^3+c$ as a quadratic expression in the variable $z^3$, etc.