# Factoring math trouble

1. Nov 15, 2013

### Husaaved

1. The problem statement, all variables and given/known data
Simplify the expression by factoring.

3(x+1)1/2(2x-3)5/2+10(x+1)3/2(2x-3)3/2

2. Relevant equations

3. The attempt at a solution

3(x+1)1/2(2x-3)5/2+10(x+1)3/2(2x-3)3/2
= (x + 1)1/2(2x-3)3/2[3(2x-3) + 10(x+1)]
= (x + 1)1/2(2x - 3)3/2(6x - 9 + 10x + 10)
= (x + 1)1/2(2x - 3)3/2(16x + 1)

This is an example problem in a textbook. What I don't understand however is what allows the first step to be a true statement, and the reasoning behind the subsequent steps. I would really appreciate some assistance with this.

Thank you.

Last edited by a moderator: Nov 15, 2013
2. Nov 15, 2013

### lurflurf

The first two steps use the distributive property
a(b+c)=ab+bc

edited to add: the rule of exponents is also used
babc=ba+c

in particular for the first step take
a=(x + 1)1/2(2x-3)3/2
b=3(2x-3)
c=10(x+1)
for the second step use the distributive property twice
first
a=3
b=2x
c=-3
second
a=10
b=x
c=1
The forth step uses the associative and commutative properties of addition
commutative
a+b=b+a
associative
a+(b+c)=(a+b)+c

now try to work through the example again

3. Nov 15, 2013

### Ray Vickson

Do you agree, or not agree, that
$$(2x-3)^{5/2} = (2x-3)^{3/2} \cdot (2x-3)^{2/2} = (2x-3)^{3/2} \cdot (2x-3)\:?$$
Do you or do you not agree that
$$(x+1)^{3/2} = (x+1)^{1/2} \cdot (x+1)^{2/2} = (x+1)^{1/2} \cdot (x+1) \:?$$
If you agree with both of these, you must then agree that there is a common factor $(2x-3)^{3/2} \, (x+1)^{1/2}$ in both terms of your original expression. So, just factor out this common thing, using the distributive law $ab + ac = a(b+c)$.