Chapter 5 - Trigonometric Identities - Section 5.2 Verifying Trigonometric Identities - 5.2 Exercises - Page 208: 28

$$4\tan^2\beta+\tan\beta-3=(\tan\beta+1)(4\tan\beta-3)$$

$$A=4\tan^2\beta+\tan\beta-3$$ We can rewrite $\tan\beta$ into $(4\tan\beta-3\tan\beta)$. In detail, $$A=4\tan^2\beta+4\tan\beta-3\tan\beta-3$$ $$A=(4\tan^2\beta+4\tan\beta)+(-3\tan\beta-3)$$ $$A=4\tan\beta(\tan\beta+1)-3(\tan\beta+1)$$ $$A=(\tan\beta+1)(4\tan\beta-3)$$