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