factorize completely x^2 2xy Y^2 3x 3y-8?
I'll assume your + key is missing, and you meant
x^2 + 2xy + y^2 + 3x + 3y - 8
(x+y)^2 + 3(x+y) - 8
Hmmm. Suppose z = x+y. Then you have
z^2+3z-8
but that does not factor with rational coefficients, so I suspect a typo. Maybe you can fix it up, and then use my hint to finish it off.
To factorize the given expression completely: x^2 + 2xy + y^2 + 3x + 3y - 8, we can follow these steps:
Step 1: Group the terms
We can group the expression by separating the quadratic terms (x^2 + 2xy + y^2) and the linear terms (3x + 3y).
(x^2 + 2xy + y^2) + (3x + 3y - 8)
Step 2: Factor the squared terms
The quadratic terms (x^2 + 2xy + y^2) can be factored as a perfect square: (x + y)^2.
(x + y)^2 + (3x + 3y - 8)
Step 3: Factor the linear terms
The linear terms (3x + 3y) have a common factor of 3. Factoring it out, we get:
3(x + y)^2 + 3(x + y) - 8
Step 4: Group common factors
Now we can see that both terms have a common factor of (x + y). We can factor it out:
3(x + y)^2 + 3(x + y) - 8 = 3(x + y)((x + y) + 1) - 8
Step 5: Simplify
Finally, we can simplify the expression further:
3(x + y)((x + y) + 1) - 8 = 3(x + y)(x + y + 1) - 8
So, the completely factorized form is 3(x + y)(x + y + 1) - 8.