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

2 Answers
May 30, 2018

Answer:

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

Explanation:

First let #t=cosx#.

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

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

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

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

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

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

So we want to add #49/4# into the expression and subtract it back out again.

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

Note that #8-49/4=32/4-49/4=-17/4#.

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

Now, note that #17/4=(sqrt17/2)^2#.

#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 #1/2# out of each part:

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

May 30, 2018

Answer:

# (cos(x) + \frac{7 + \sqrt(17)}{2})(cos(x) + \frac{7 - \sqrt(17)}{2})#

Explanation:

let # u = cos(x) #
The question then becomes:

Factor #u^2+7u+8# you could just use quadratic formula here i.e. # u = \frac{-b\pm \sqrt(b^2-4ac)}{2a} #

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, # r_1# and #r_2 # such that # (u-r_1)(u - r_2) = u^2+7u+8#

Expand: # (u-r_1)(u - r_2) = u^2 - r_1u - r_2u + (r_1)(r_2)#
# = u^2 - (r_1+r_2)u + (r_1)(r_2) #

Thus: # u^2 - (r_1+r_2)u + (r_1)(r_2) = u^2+7u+8 #
and therefore: # - (r_1+r_2) = 7# and #(r_1)(r_2) = 8 #

# (r_1+r_2) = -7, (r_1+r_2)^2 = 49 #
# (r_1)^2 + 2(r_1)(r_2) + (r_2)^2 = 49 #
# (r_1)^2 + 2(r_1)(r_2) + (r_2)^2 - 4(r_1)(r_2) = 49 - 4(8) = 17 #
# (r_1)^2 - 2(r_1)(r_2) + (r_2)^2 = 17 #

# (r_1-r_2)^2 = 17 #
# r_1-r_2 = \sqrt(17) #
# \frac{r_1+r_2 + r_1-r_2}{2} = r_1 = \frac{-7 + \sqrt(17)}{2} #
# \frac{r_1+r_2 - (r_1-r_2)}{2} = r_2 = \frac{-7 - \sqrt(17)}{2} #

Thus, the factored form is # (u + \frac{7 + \sqrt(17)}{2})(u + \frac{7 - \sqrt(17)}{2})#

sub # u = cos(x)# to get:

# (cos(x) + \frac{7 + \sqrt(17)}{2})(cos(x) + \frac{7 - \sqrt(17)}{2})#