Question

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

Answer 1

`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)`

Answer 2

`(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})`