How do you factor cos2x+7cosx+8?

2 Answers
mason m
May 30, 2018

14(2cosx+7+√17)(2cosx+7−√17)

Explanation:

First let t=cosx.

y=t2+7t+8

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

y=(t2+7t)+8

Note that (t+72)2=(t+72)(t+72)

=t2+72t+72t+(72)2

=t2+7t+494

So we want to add 494 into the expression and subtract it back out again.

y=(t2+7t+494)+8−494

Note that 8−494=324−494=−174.

y=(t+72)2−174

Now, note that 174=(√172)2.

y=(t+72)2−(√172)2

Now, we have a difference of squares and can factor it as one.

y=[(t+72)+√172][(t+72)−√172]

y=(cosx+7+√172)(cosx+7−√172)

If we wish, we can bring a common factor of 12 out of each part:

y=14(2cosx+7+√17)(2cosx+7−√17)

Roy D.
May 30, 2018

(cos(x)+7+√172)(cos(x)+7−√172)

Explanation:

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

Factor u2+7u+8 you could just use quadratic formula here i.e. u=−b±√b2−4ac2a

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, r1 and r2 such that (u−r1)(u−r2)=u2+7u+8

Expand: (u−r1)(u−r2)=u2−r1u−r2u+(r1)(r2)
=u2−(r1+r2)u+(r1)(r2)

Thus: u2−(r1+r2)u+(r1)(r2)=u2+7u+8
and therefore: −(r1+r2)=7 and (r1)(r2)=8

(r1+r2)=−7,(r1+r2)2=49
(r1)2+2(r1)(r2)+(r2)2=49
(r1)2+2(r1)(r2)+(r2)2−4(r1)(r2)=49−4(8)=17
(r1)2−2(r1)(r2)+(r2)2=17

(r1−r2)2=17
r1−r2=√17
r1+r2+r1−r22=r1=−7+√172
r1+r2−(r1−r2)2=r2=−7−√172

Thus, the factored form is (u+7+√172)(u+7−√172)

sub u=cos(x) to get:

(cos(x)+7+√172)(cos(x)+7−√172)

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