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