Question

Factorise (2xy-2xz-y+2)

Answer 1

To factorise the expression \(2xy - 2xz - y + 2\), we can look for common factors in pairs of terms.

First, we can group the terms:

\[
(2xy - 2xz) + (-y + 2)
\]

Now, we can factor out the common factors in each group:

1. In the first group \(2xy - 2xz\), we can factor out \(2x\):

\[
2x(y - z)
\]

2. In the second group \(-y + 2\), we can factor out \(-1\):

\[
-1(y - 2)
\]

Now rewriting the entire expression gives us:

\[
2x(y - z) - 1(y - 2)
\]

This does not seem to lead to a common factor anymore, but we can check if there is a way to factor this expression differently.

We observe that we can rearrange the grouping:

\[
2xy - y - 2xz + 2
\]

Now we can refactor by grouping again:

1. From \(2xy - y\), factor out \(y\):

\[
y(2x - 1)
\]

2. From \(-2xz + 2\), factor out \(-2\):

\[
-2(z - 1)
\]

Putting it all together results in:

\[
y(2x - 1) - 2(z - 1)
\]

Now, let's check if we can factor by rearranging terms:

To factor the original expression \(2xy - 2xz - y + 2\):

\[
= 2(xy - xz) - (y - 2)
\]

This gives us a way to factor as follows:

\[
= 2x(y - z) - (y - 2)
\]

The final factored form is:

\[
2x(y - z) - (y - 2)
\]

It does not appear to factor further simply, but this rearrangement and understanding of grouping yields a structured form ready for any substitutions or further manipulations desired.

However, we could also state it in different reorganizations depending on what is deemed 'factored' or acceptable based on your needs.

If factoring was meant in a simpler polynomial context, we may not achieve much more than what we have done:

\[
= 2x(y - z) - 1(y - 2)
\]

So the simplest form we can group the terms would be:

\[
= (2x - 1) (y - z) + 2
\]

Or:

\[
2xy - 2xz - y + 2 = (y - 2) + 2x(y - z)
\]

However, thus described as above does not yield a common polynomial factoring.

Thus, the conclusion is **Group-Based Factorisation**:

### \(2x(y - z) - (y - 2)\)