1. GCF factoring Question

Regarding GCF factoring, can the same answer be derived from multiple number sets? Specifically fin the following problem:

$\displaystyle 40a^5x+80a^5y-120a^5z$ & its answer 40a^5(x + 2y - 3z)

40 obviously is the GCF and goes into all three numbers. But what if one doesn't realize this initially and decides to approach it as such?

pulling out the terms
$\displaystyle 40a^5x = 2^2*10$
$\displaystyle 80a^5y=2^4*5$
$\displaystyle -120a^5z= 2^3*15$

then finding the lowest power of each variable and number. Will one still arrive at the same answer as if they used 40 as the GCF?

2. This is true.

$\displaystyle 40=2^3\cdot5$

$\displaystyle 80=2^4\cdot5$

$\displaystyle 120=2^3\cdot3\cdot5$

Use all the common factors with the least exponent.

Thus, we have $\displaystyle 2^3\cdot5=40$

Do the same for the variables. Find the common variables and use the least exponent.

3. Yikes, you changed yours up a little bit in your example

Instead of what I used, 2^2 * 10 = 40 for the first term you used 2^3 * 5 was. So is answer is already wrong at this point, going with 2^2 * 10 = 40? If so how can I avoid future problems such as this?

4. Originally Posted by cmf0106
Yikes, you changed yours up a little bit in your example

Instead of what I used, 2^2 * 10 = 40 for the first term you used 2^3 * 5 was. So is answer is already wrong at this point, going with 2^2 * 10 = 40? If so how can I avoid future problems such as this?

You should factor your numbers into the product of primes.

$\displaystyle 40=2^3\cdot5$, otherwise you may miss an exponent. You were lucky this time since you already had a $\displaystyle 2^3$ as a factor of 120.

5. Originally Posted by cmf0106
Yikes, you changed yours up a little bit in your example

Instead of what I used, 2^2 * 10 = 40 for the first term you used 2^3 * 5 was. So is answer is already wrong at this point, going with 2^2 * 10 = 40? If so how can I avoid future problems such as this?
When factoring numbers to find the GCF, you should factor everything completely into powers of prime numbers. 10 is not prime, because $\displaystyle 10=2\cdot5$, but 2 and 5 are both prime. Once you have factored completely, then you can take the common factors, and their product will be your greatest common factor.

6. Could not have said it better myself.