# Factoring Help

• Sep 8th 2007, 05:55 PM
chris123
Factoring Help
I need some help on some problems.

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

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

3.
$x^2+2
$

______
$x
$

4.
$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
• Sep 8th 2007, 11:42 PM
red_dog
1. $25-10a+a^2-y^2=(5-a)^2-y^2$ and use $a^2-b^2=(a-b)(a+b)$

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

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

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

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

7. $\displaystyle\frac{x}{x+2}=\frac{3x}{x-1}\Leftrightarrow x(x-1)=3x(x+2)$ and you continue.
• Sep 9th 2007, 11:03 AM
chris123
Thanks for the help, could you help me on these aswell or tell me the method to get the answer?

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

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

3.
$
\frac{1}{x}+\frac{1}{y}
$

_____
$
\frac{1}{x}-\frac{1}{y}
$

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

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

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

_________
$
2+\frac{1}{x+2}
$

7. $
\frac{6x}{(x+2)(x+3)}-\frac{5}{(x+2)(x-3)}
$
• Sep 9th 2007, 03:29 PM
Soroban
Hello, chris123!

Here is some help . . .

Quote:

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

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

Quote:

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

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

Quote:

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

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

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

$