1. ## Basic Factoring Question

Hello I am just starting factoring, the book lists the correct method for solving the following problem as.
$8x+8= 8*x+8*1=8(x+1)$

but what if you were to do it like this?
$8x+8=4*2x + 4*2=4(2x+2)$

Very confusing since they both give the correct answer once distributed.

2. Originally Posted by cmf0106
Hello I am just starting factoring, the book lists the correct method for solving the following problem as.
$8x+8= 8*x+8*1=8(x+1)$

but what if you were to do it like this?
$8x+8=4*2x + 4*2=4(2x+2)$

Very confusing since they both give the correct answer once distributed.
the last is not fully factored, since there is still a common factor of 2 among the terms in the brackets. so yes, it is still correct, but as far as factoring goes, it is incomplete

3. Thanks & one last thing for now.
In $12xyz(4x^4y^2z^5+5x^3z^2+3x^5y)$
$x^2$ can be factored out because in the term $5x^3z^2$ $x^2$ is the MOST that can removed and still remain an x value. And since factoring is "divided into" we are subtracting the exponent values because $\frac{a^m}{a^n}=a^{m-n}$. Thus $x^2$ can be factored out, does that logic sound correct? Also this procedure is not done for the z and y variables because they do not show up in all the terms right?

4. Originally Posted by cmf0106
Thanks & one last thing for now.
In $12xyz(4x^4y^2z^5+5x^3z^2+3x^5y)$
$x^2$ can be factored out because in the term $5x^3z^2$ $x^2$ is the MOST that can removed and still remain an x value. And since factoring is "divided into" we are subtracting the exponent values because $\frac{a^m}{a^n}=a^{m-n}$. Thus $x^2$ can be factored out, does that logic sound correct? Also this procedure is not done for the z and y variables because they do not show up in all the terms right?
basically, you can factor out the lowest power of any term common to all terms. in the case of $x$, it is $x^3$

your logic seems about right, but i would call it "dividing out of" but maybe that makes no sense. you are correct with y and z. but yes, you watch the powers in the same way you stated. and you can double check yourself by seeing that when you are multiplying out again, the powers add up to give you what was in the original

5. Same type of question for this one $28xy^2-14x$
The book has the answer $-7x(-4y^2+2$ but I solved it
14x * $2y^2$ - 14x * 1= $14x(2y^2-1)$. It seems like I picked the correct number combination, because in the original problem 14x is the largest number that divides into both 14x and 28xy

6. Originally Posted by cmf0106
Same type of question for this one $28xy^2-14x$
The book has the answer $-7x(-4y^2+2$ but I solved it
14x * $2y^2$ - 14x * 1= $14x(2y^2-1)$. It seems like I picked the correct number combination, because in the original problem 14x is the largest number that divides into both 14x and 28xy
your answer is better. good job. the book still left a factor of 2 to be taken out