factorise completely:2ax-4bx-ay-2by
2ax - 4bx - ay - 2by.
2x(a-2b) - y(a-2b).
(a-2b)(2x-y).
2ax-4bx-ay-2by
= (2ax-4bx) + (-ay-2by)
= (2*a*x-2*2*b*x) + (-1*a*y-1*2*b*y)
= 2x(a-2b) - y(a+2b)
=2x(a-2b) +y (a-2b)
= (a-2b) (2x+y)
Excellent
Well, let's call this expression a math joke. It wants to be factorized completely, just like a clown wants to squeeze into a tiny car! So let's give it a shot:
First, let's take out the greatest common factor from each term. This will involve finding the common factors of the coefficients and the variables.
The coefficients 2, -4, -1, and -2 all have a common factor of 2. The variables a and b are both present in each term.
So, let's factor out 2ab:
2ax - 4bx - ay - 2by = 2ab(x - 2b) - a(y + 2b)
And, voila! We've successfully factorized this expression completely. It's like watching a clown pop out of a tiny car – surprising, yet amusing!
To factorize the given expression completely, we need to find the common factors in each term. Let's analyze each term separately:
2ax: The common factor here is 2a.
-4bx: The common factor here is -4b.
-ay: The common factor here is -a.
-2by: The common factor here is -2b.
Now, let's factor out these common factors:
2ax - 4bx - ay - 2by = 2a(x - 2b) - a(y + 2b)
Notice that we can still factor out -a from the second term:
2a(x - 2b) - a(y + 2b) = 2a(x - 2b) - a(2b + y)
Now, we can factor out the common factor of 2a from both terms:
2a(x - 2b) - a(2b + y) = 2a(x - 2b) - 2a(b + y)
Finally, we can factor out the remaining common factor of 2a:
2a(x - 2b) - 2a(b + y) = 2a[(x - 2b) - (b + y)]
Simplifying further, we get:
2a[(x - 2b) - (b + y)] = 2a(x - 2b - b - y)
Combining like terms, we obtain the fully factorized expression:
2a(x - 3b - y)