how do you factor cot^4x+2cot^2x+1?

Question

campbell_st · Answer

its an equation reducible to a quadratic, think of it as 

$$(\cot^2(x))^2 + 2\cos^2(x)+ 1$$\

so it looks like a perfect square

ajay1010 · Answer

?

whpalmer4 · Answer

let $ u = \cot(x)$
then rewrite as $$u^4 + 2u^2+1$$
can you factor that?

ajay1010 · Answer

i believe so?

campbell_st · Answer

just make a simple substitution 

$$u = \cos^2(x)$$

then 

$$(\cot^2(x)^2 + 2\cot^2(x) + 1 = u^2 + 2u + 1$$

whpalmer4 · Answer

and you could do another level of substitution:
$$z = u^2$$$$z^2 + 2z + 1$$which I am certain you can factor.  Then undo the substitutions in your factored version and you are done.

campbell_st · Answer

then if you factor you get 

$$u^2 + 2u + 1 = (u + 1)^2$$

now make the reverse substitution for u

hope it helps

whpalmer4 · Answer

@campbell_st gets it done in one step, but that might be a bit hard to grasp if you aren't used to doing this.

ajay1010 · Answer

im not used to any of this> first time

whpalmer4 · Answer

but in general, if you have something complicated, see if you can reduce it to something simpler to make the algebra easier, then expand it out when you have the answer.  Pretty common trick...

campbell_st · Answer

ok... so can you factor quadratics... as this is the key to this question... 

recognising the powers form a quadratic equation with the correct substituion

campbell_st · Answer

the key to this question is recongising the powers    4, 2 and 0

the same idea can be applied if the powers where 6, 3 , 0

of 10, 5, 0