# Hi, could anyone help me with a algebra question

1. May 6, 2007

### ihopeican

hi there,

i just forgot how to solve quadratics and was wondering if anyone could help me on a question it is quite easy.

(x+3)(x-2)^2= 72
(x-2)^2
=x^2-4x+4
(x+3)(x^2-4x+4)
x^3-4x^2+4x+3x^2-12x+12

=x^3-x^2-8x+12= 72

im not quite sure what i should do next if i havent made any mistakes so far.

thankyou,

2. May 6, 2007

### Office_Shredder

Staff Emeritus
I would start by noting 72=9*8=3*3*8.

This should get you one solution.

Then when you get to

x^3-x^2-8x+12= 72

Subtract to get

x3-x2-8x-60=0

and factor out x-a, where a is the solution you found earlier.

3. May 7, 2007

### turdferguson

Guess factors of 60 to plug in using synthetic division/substitution

4. May 8, 2007

### light_bulb

multiply each term of one binomial by each term of of the other

(x + 3)(x-2)^2=72

multiply each term by 3
3 * -2 = -6
3 * x = 3x

combine like terms
x^2 + x3 - x2 -6

is

x^2 + x -6

now

x^2 + x -6 + 72

do the quadratic eq and check for 0