Factoring Trigonometric Expressions...3

[nycmathguy]

Section 5.1

Question 30

6 cos^2 x + 5 cos x - 6

Let u = cos x

6u^2 + 5u - 6

Factor by grouping.

6(-6) = -36

9(-4) = -36

9 + (-4) = 5

6u^2 + 9u - 4u - 6

6u^2 + 9u becomes 3u(2u + 3)

-4u - 6 becomes -2(2u + 3)

We now have (3u - 2)(2u + 3).

Back-substitute for u.

(3cos x - 2)(2cos x + 3)

You say?

 

[MathLover1]

correct

 

[nycmathguy]

correct

Thanks.