# Factor this expression

1. Mar 27, 2012

### cupcakes

1. The problem statement, all variables and given/known data
Express the following as A2+B2:

3[√3+√5+√7]2

2. Relevant equations

3. The attempt at a solution

I expanded it to 45 + 6√15 + 6√21 + 6√35
Should I collect like terms (the multiples of 6)? I don't know how to proceed from here. Thanks in advance for any help.

2. Mar 27, 2012

### tal444

I'm not sure, but I don't think this expression can be converted to the form A2+B2.

3. Mar 27, 2012

### cupcakes

Maybe it's because I forgot to mention that A & B may contain roots. I think it's possible because it is an assigned question. But I'm stuck...

4. Mar 27, 2012

### e^(i Pi)+1=0

There are no real numbers that factor to the sum of two squares. Your answer will be imaginary.

5. Mar 28, 2012

### chubbyorphan

3[√3+√5+√7]^2

3[√(1.5)+√(2.5)+√(3.5)]^2 + 3[√(1.5)+√(2.5)+√(3.5)]^2

up in the middle of the night doing this.. so I may have broken a million rules getting to this point
You Might want to double check but they seem equivalent

6. Mar 28, 2012

### Staff: Mentor

3[√3+√5+√7]2 = 2[√3+√5+√7]2 + [√3+√5+√7]2

The original expression is now written as a sum of two terms. Can you finish the problem by showing that each of these terms is the square of something? I.e., can you identify A and B with the above being equal to A2 + B2?

7. Mar 28, 2012

### chubbyorphan

It only works with odd numbers

(√1.5 + √3.5)2 + (√1.5 + √3.5)2 = (√3 + √7)2

The left side is 2 * 9.58 = 19.16. The right side is 19.16

(√44.5 + √6.5)2 + (√44.5 + √6.5)2 = (√89 + √13)2

The left side is 2 * 85.014 = 170.029. The right side is 170.029

yesh?

Last edited by a moderator: Mar 28, 2012
8. Mar 28, 2012

### Staff: Mentor

Yes, this works. My example (now deleted) was flawed in that I forgot to square the value on the right side. Apologies for the misdirection...

Here's what's going on.
(√1.5 + √3.5)2 + (√1.5 + √3.5)2 = 2(√1.5 + √3.5)2
= (√2 *√1.5 + √2 *√3.5)2
= (√3 + √7)2

9. Mar 29, 2012

### chubbyorphan

Hey no worries, Mark44, beside I found my version of the solution completely out of luck just bored and messing around with my calculator :P
You're version of the sum of two terms is just as valid isn't it?

10. Apr 4, 2012

### cupcakes

How did I not notice that?! :rofl: That works perfectly Mark. Thanks to everyone for their help.