1. ## Factoring Help

I need some help on some problems.

1. $\displaystyle 25-10a+a^2-y^2$

2.
$\displaystyle y^2-4x^2-20x-25$

3.
$\displaystyle x^2+2$
______
$\displaystyle x$

4.
$\displaystyle 4y(y-3)^3(y+4)^4+6y(y-3)^5(y+4)^2$

5.
1.. 1
- + -
x.. 2

6.
3x.. 1.. 1
-- + - = -x - 2
4... 2.. 3

7.
.x......3x
---. = ---
x+2....x-1

2. 1. $\displaystyle 25-10a+a^2-y^2=(5-a)^2-y^2$ and use $\displaystyle a^2-b^2=(a-b)(a+b)$

2. $\displaystyle y^2-4x^2-20x-25=y^2-(2x+5)^2$ and use $\displaystyle a^2-b^2=(a-b)(a+b)$

4. $\displaystyle 4y(y-3)^3(y+4)^4+6y(y-3)^5(y+4)^2=2y(y-3)^3(y+4)^2[2(y+4)^2+3(y-3)^2]$

5. $\displaystyle \displaystyle\frac{1}{x}+\frac{1}{2}=\frac{x+2}{2x }$

6. $\displaystyle \displaystyle \frac{3x}{4}+\frac{1}{2}=\frac{x}{3}-2$
Multiply both sides by 12.

7. $\displaystyle \displaystyle\frac{x}{x+2}=\frac{3x}{x-1}\Leftrightarrow x(x-1)=3x(x+2)$ and you continue.

3. Thanks for the help, could you help me on these aswell or tell me the method to get the answer?

1. $\displaystyle \frac{1}{x}+\frac{1}{y}+\frac{1}{w}$

2. $\displaystyle 5-\frac{3}{x}$

3.
$\displaystyle \frac{1}{x}+\frac{1}{y}$
_____
$\displaystyle \frac{1}{x}-\frac{1}{y}$

4. $\displaystyle \frac{2}{x+1}+\frac{3}{x-1}$

5. $\displaystyle \frac{1}{x+2}+5$

6.
$\displaystyle \frac{1}{x+2}+3$
_________
$\displaystyle 2+\frac{1}{x+2}$

7. $\displaystyle \frac{6x}{(x+2)(x+3)}-\frac{5}{(x+2)(x-3)}$

4. Hello, chris123!

Here is some help . . .

$\displaystyle 3)\;\;\frac{\frac{1}{x}+\frac{1}{y}}{\frac{1}{x}-\frac{1}{y}}$
Multiply top and bottom by $\displaystyle xy$:

. . $\displaystyle \frac{xy\left(\frac{1}{x}+\frac{1}{y}\right)}{xy\l eft(\frac{1}{x} - \frac{1}{y}\right)} \;=\;\frac{y + x}{y - x}$

$\displaystyle 6)\;\;\frac{\frac{1}{x+2}+3}{2+\frac{1}{x+2}}$
Multiply top and bottom by $\displaystyle (x+2)$

. . $\displaystyle \frac{(x+2)\left(\frac{1}{x+2} + 3\right)}{(x+2)\left(2 + \frac{1}{x+2}\right)} \;=\;\frac{1+3(x+2)}{2(x+2) + 1} \;=\;\frac{1 + 3x +6}{2x + 4+ 1} \;=\;\frac{3x+7}{2x+5}$

$\displaystyle 7)\;\;\frac{6x}{(x+2)(x+3)}-\frac{5}{(x+2)(x-3)}$
Get a common denominator . . .

$\displaystyle \frac{6x}{(x+2)(x+3)}\cdot{\color{blue}\frac{x-3}{x-3}} \;- \;\frac{5}{(x+2)(x-3)}\cdot{\color{blue}\frac{x+3}{x+3}} \;\;= \;\;\frac{6x(x-3) \,- \,5(x+3)}{(x+2)(x+3)(x-3)}$

. . $\displaystyle = \;\frac{6x^2-18x-5x-15}{(x+2)(x+3)(x-3)} \;\;=\;\;\frac{6x^2-23x - 15}{(x+2)(x+3)(x-3)}$