Could you explain how to factor completely
-a^3n+1 + 2b^2n+b - ab - (2a^3n*b^2n)
Sure! To factor completely, we need to look for common factors and then use various factoring techniques such as factoring by grouping, difference of squares, or perfect square trinomials.
Let's start by examining the given expression:
-a^3n+1 + 2b^2n+b - ab - (2a^3n*b^2n)
Step 1: Look for common factors.
In this expression, we can see that each term has a common factor of (-1). Let's factor it out:
(-1) * (a^3n-1) + (-1) * (2b^2n+b) + (-1) * (ab) - (-1) * (2a^3n*b^2n)
Now, let's simplify it further:
(-1)(a^3n-1) - (2b^2n+b) - ab + (2a^3n*b^2n)
Step 2: Group the terms and factor out common factors.
Group the terms in pairs and factor out any common factors for each pair:
[(-1)(a^3n-1) - ab] + [(2a^3n*b^2n) - (2b^2n+b)]
Now, let's factor out the common terms:
-a(a^3n-1) - b(a^3n-1) + 2b^2n(a^3n-1) - b(a^3n-1)
Step 3: Look for common factors again.
This time, we can see that each pair of terms has a common factor of (a^3n-1). Let's factor it out:
(a^3n-1)(-a - b + 2b^2n - b)
Simplifying it further, we have:
(a^3n-1)(-a - 2b + 2b^2n)
Therefore, the expression -a^3n+1 + 2b^2n+b - ab - (2a^3n*b^2n) can be factored completely as (a^3n-1)(-a - 2b + 2b^2n).