Need help with this pre-calculus question?

Question

3 (x+2)^2 (x-3)^2 - (x+2)^3 (2) (x-3)

---OVER---

(x-3)^4

I can't figure out how to go about this problem, and the book is focusing more on rationalizing numerators or denominators more so than simplifying expressions. Any useful input will help.

Thanks!

Answer 1

3 (x+2)^2 (x-3)^2

---OVER---

(x-3)^4

Cancel (x-3)

= 3(x+2)²/(x-3)²

(x+2)^3 (2) (x-3)

---OVER---

(x-3)^4

= (x+2)^3 (2)

---OVER---

(x-3)³

= 2(x+2)³/(x-3)³

Combine those two becomes

3(x+2)²/(x-3)² - 2(x+2)³/(x-3)³

common factor is (x+2)²

= (x+2)² [3/(x-3)² - 2(x+2)/(x-3)³]  -----------------(1)

3/(x-3)² = 3(x-3)/(x-3)³ = (3x-9)/(x-3)³

2(x+2)/(x-3)³ = (2x+4)/(x-3)³

[3/(x-3)² - 2(x+2)/(x-3)³]

= [(3x-9)/(x-3)³ - (2x+4)/(x-3)³ ]

= (x-13)/(x-3)³

(1) becomes (x+2)²(x-13)/(x-3)³