# Factorising

• July 11th 2008, 05:17 PM
princess_anna57
Factorising
solve (x-5)(x+2) = 3x+2

Here's what I got so far:

x^2 + 2x - 5x - 10 = 3x + 2
x^2 - 3x - 10 = 3x + 2
x^2 - 6x - 12 = 0

and then I don't know what to do next. Do you do (x ) (x )? I don't know how to figure out how to get the missing numbers.

I would also like to know how to solve these two:
5x = -2x^3
5x = 2x^3
• July 11th 2008, 05:53 PM
Reckoner
Quote:

Originally Posted by princess_anna57
solve (x-5)(x+2) = 3x+2

Here's what I got so far:

x^2 + 2x - 5x - 10 = 3x + 2
x^2 - 3x - 10 = 3x + 2
x^2 - 6x - 12 = 0

and then I don't know what to do next. Do you do (x ) (x )? I don't know how to figure out how to get the missing numbers.

Complete the square or use the quadratic formula. You won't get nice factors with integer coefficients for this one.

Quote:

Originally Posted by princess_anna57
I would also like to know how to solve these two:
5x = -2x^3
5x = 2x^3

This one should be easy.

$5x = -2x^3\Rightarrow 2x^3 + 5x = 0\Rightarrow x(2x^2 + 5) = 0$

Only one of the solutions will be real.