Right off the bat, we should notice that each term has an #x# in it. That means we can take an #x# out:
#10x^3-53x^2+70x=x(10x^2-53x+70)#
Now it gets a little tricky. We have the trinomial #10x^2-53x+70#, and we can't simplify it any further. So we'll have work with what we have.
Factoring something like this is hard because we have that #10x^2# term. If there was no #10#, it would be considerably easier to factor. To work around this, we'll use the AC method of factoring.
The first step in this method is to multiply the coefficient (the number in front of the #x#) of the #x^2# term by the constant term (the term without an #x#). That means multiplying #10# and #70#: #10*70=700#.
Next step is finding two numbers that multiply to #700# and add to the coefficient on the #x# term (in this case, #-53#). Important to note here is that these two numbers must be negative, because their sum needs to be negative (this won't be an issue, since a negative times a negative is positive - and these two numbers must also multiply to positive #700#).
Ok, first find two numbers that add to #-53# and multiply to something close to #700#:
#-30*-23=690#
#-29*-24=696#
#-28*-25=700#
We found our magic numbers: #-28# and #-25#. Now we rewrite #10x^2-53x+70# as #10x^2-25x-28x+70#. Factor out a #5x# from #10x^2-25x# and a #-14# from #-28x+70#:
#10x^2-53x+70=5x(2x-5)-14(2x-5)#
#2x-5# is a common term, so we can pull it out to get:
#10x^2-53x+70=(2x-5)(5x-14)#
So #10x^2-53x+70# factors into #(2x-5)(5x-14)#. Putting this back into the first equation, we have our final result:
#10x^3-53x^2+70x=x(2x-5)(5x-14)#
Yes, I know this is a lot of work and is difficult to understand, but with practice you should get this method down.
