Question 336901
it's ok! there's always time to fix your errors and re learn what you already know

firstly, a brief explanation about factoring. factoring is simply the process of breaking stuff down into components thats can't be broken down further. let's start with a simple example, take the number 6
6=3*2, now we can't break 2 and 3 further since they are prime. hence we say, 6's factors include 2 and 3

in the same sense the polynomial we have may jus be composed of other parts that cannot be broken down

2(a^2)(x^2)-6a^2x+4ax
so we can notice there are leading coeffecients for each term, namely, 2, -6, and 4
also there are 'a' terms and ther are also 'x' terms. first let's factor out the leading coeffecients, what factor do 2,-6 and 4 have in common? it's 2 of course! since 2*1=2,2*-3=6 and 2*2=4

so we can factor out 2 from each term and get

2(a^2)(x^2)-6a^2x+4ax=2[(a^2)(x^2)-3a^2x+2ax] <- expand and check to make sure this is true

next we look at 'a' terms, and see that we have a^2 in the first term, a^2 in the second and a in the third, therefore the common factor for them will be just a, so let's pull that out as well

2a[(a)(x^2)-3ax+2x]

lastly we do the same thing for 'x' terms. i will leave it to u to show the common factor is x

2ax[(a)(x)-3a+2] <- factored form, we cannot simplify anymore