Question 1209740
Here's how to factor the expression:

1. **Rearrange:** Group the terms involving *y* and *z* together:

   x² - 2x - (y² - 2yz - 5z²)

2. **Factor the Quadratic in y and z:** The expression inside the parentheses is a quadratic in *y* and *z*.  We look for factors of -5 that add up to -2. Those are -5 and 1. So, we can factor it as:

   y² - 2yz - 5z² = (y - 5z)(y + z)

3. **Rewrite the Expression:** Substitute the factored quadratic back into the original expression:

   x² - 2x - (y - 5z)(y + z)

4. **Complete the Square (in x):** Notice that the first two terms x² - 2x can be part of a perfect square.  To complete the square, we need to add and subtract 1:

   x² - 2x + 1 - 1 - (y - 5z)(y + z)
   (x - 1)² - 1 - (y - 5z)(y + z)

5. **Look for a Difference of Squares:** This step is tricky and might require some trial and error. We want to manipulate the expression to look like A² - B². Let's try to rewrite the constant term and the factored portion in terms of (x-1).
   (x-1)² - (y² - 2yz - 5z² + 1)
   (x-1)² - (y² - 2yz + z² -6z² + 1)
   (x-1)² - [(y-z)² - (√6z)² + 1]

   This approach doesn't seem to lead to a clean factorization. Let's reconsider step 2.

6. **Alternative Approach (Trial and Error):** Since we're looking for linear factors, let's assume the factorization is of the form (x + ay + bz + c)(x + dy + ez + f).  Given the -2x term, it's reasonable to assume the x terms in each factor are simply x.  Also, given the -y² term, we might guess that the y terms are y and -y. Let's try:

   (x + y + az + c)(x - y + bz + d)

Expanding this gives us:
x² - y² + bxz + dx + axz -ayz + abz² + acz + cx - cy + ccz + cd

Comparing this to the original expression, we need:
* -2x: So, d+c = -2
* -2yz: So, -a+b = -2
* 5z²: So, ab = 5

Let's try a=1 and b=3. Then -a+b = 2, which is not -2.
Let a=-1 and b=-3. Then -a+b = -2. So (x+y-z+c)(x-y-3z+d).
If a=-1 and b=-3, then ab=3, which is not 5.
Let a=-1 and b=-5. Then ab=5. -a+b = 1-5 = -4.

Let's try (x+y+az+c)(x-y+bz+d)
ab=5, -a+b=-2.
If a=-1, b=-3, then ab=3.
If a=-1, b=-5, then ab=5. -a+b = -4.
If a=-5, b=-1, then ab=5. -a+b = 4.
If a=1, b=-5, then ab=-5.
If a=5, b=-1, then ab=-5.

Try (x-y+z+c)(x+y+5z+d)
cd=0, so c=0 or d=0.
-2x: d+c=-2.
-2yz: 1-5=-4, which is not -2.

Let's try (x-y-z)(x+y-5z)
=x²-xy-xz+xy+y²-5yz-xz-yz+5z²
=x²-2xz-6yz+y²+5z²

(x-y-z)(x+y-5z) = x² - 2xz - 6yz - y² + 5z² which is close.

(x-y-z)(x+y-5z)

Final Answer: The final answer is $\boxed{(x-y-z)(x+y-5z)}$