# How do you factor #cos^2 x+7 cos x+8#?

##### 2 Answers

#### Explanation:

First let

#y=t^2+7t+8#

Now, let's complete the square to factor this.

#y=(t^2+7t)+8#

Note that

#=t^2+7/2t+7/2t+(7/2)^2#

#=t^2+7t+49/4#

So we want to add

#y=(t^2+7t+49/4)+8-49/4#

Note that

#y=(t+7/2)^2-17/4#

Now, note that

#y=(t+7/2)^2-(sqrt17/2)^2#

Now, we have a difference of squares and can factor it as one.

#y=[(t+7/2)+sqrt17/2][(t+7/2)-sqrt17/2]#

#y=(cosx+(7+sqrt17)/2)(cosx+(7-sqrt17)/2)#

If we wish, we can bring a common factor of

#y=1/4(2cosx+7+sqrt17)(2cosx+7-sqrt17)#

#### Explanation:

let

The question then becomes:

Factor

or you could do it the long way (which isn't any better than the formula, in fact it's one of the methods use to formulate the quadratic formula) :

find two roots,

Expand:

Thus:

and therefore:

Thus, the factored form is

sub