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

2 Answers
May 30, 2018

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

(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})