Originally Posted by

**Leah<3** 1- divide using long division (9x^3-9x^2-x-12) / (3x-5)

2- simplify 3/x^2-9 + 4/x-3

3- given the polynomial x^3-5x^2+2x+8 where x-4 is a factor. find another factor

4- simplify 4x^5y^-3 / 20x^-2y

...

Hello,

to #1: Code:

(9x³ - 9x² - x - 12) ÷ (3x - 5) = 3x² + 2x + 3
(3x³ - 15x²)
-------------
6x² - x
6x² - 10x
---------
9x - 12
9x - 15
--------
3 ← remainder

to #2:

$\displaystyle \frac{3}{x^2-9} + \frac{4}{ x-3}=\frac{3}{ (x+3)(x-3)}+\frac{4(x+3)}{(x+3)(x-3)} = \frac{4x+15}{x^2-9}$

to #3:

a) use long division

b) use synthetic division Code:

| 1 -5 2 8
| 0 4 -4 -8
------|----------------
x = 4 | 1 -1 -2 0

That means: $\displaystyle x^3 -5x^2 + 2x + 8 = (x-4)(x^2 - x - 2)$

The last factor can be factorized again so that you finally have:

$\displaystyle x^3 -5x^2 + 2x + 8 = (x-4)(x- 2)(x+1)$

to #4:

$\displaystyle \frac{4x^5 y^{-3}}{20x^{-2} y} = \frac{1}{5} x^{5-(-2)} y^{-3 - 1} = \frac{1}{5} x^7 y^{-4} = \frac{x^7}{5y^4}$