SOLUTION: Kuta Software Infinite Algebra 2: The Remainder Theorem ... - Free Printable

Okay, now I see the second problem on the right. The expression to simplify is $3x^2 - 2x + 4$. Below it, there's some work that appears to involve factoring or grouping, but it's very blurry. I can make out what looks like $(3x^2 - 2x)$ and then $+4$, which suggests an attempt to group terms. There also seems to be a step involving $-2x$ and possibly a factor of 2. However, the text is too indistinct to follow the logic accurately. Since the task is to "Simplify each expression," and both expressions are quadratic, I will now focus on solving the first problem correctly, as the provided work is unclear.

I will now analyze the first expression, $x^2 - y^2 - 2y$, to provide a correct solution.

The expression is $x^2 - y^2 - 2y$.
This expression contains three terms: $x^2$, $-y^2$, and $-2y$.
The first two terms, $x^2 - y^2$, form a difference of squares, which can be factored as $(x - y)(x + y)$.
However, the presence of the $-2y$ term means we cannot simply factor the entire expression as a difference of squares.
We can try to group the terms differently. Let's group the y-terms together:
$x^2 - (y^2 + 2y)$
Now, we can complete the square for the expression inside the parentheses, $y^2 + 2y$.
To complete the square, we take half of the coefficient of y, which is 2, so half is 1, and then square it to get 1.
So, $y^2 + 2y = (y^2 + 2y + 1) - 1 = (y + 1)^2 - 1$.
Substituting this back into the original expression, we get:
$x^2 - ((y + 1)^2 - 1)$
Distributing the negative sign, we get:
$x^2 - (y + 1)^2 + 1$
Now, we have $x^2 - (y + 1)^2$, which is again a difference of squares.
Let $a = x$ and $b = y + 1$, so $a^2 - b^2 = (a - b)(a + b)$.
Therefore, $x^2 - (y + 1)^2 = (x - (y + 1))(x + (y + 1)) = (x - y - 1)(x + y + 1)$.
Adding the +1 from earlier, the simplified expression is $(x - y - 1)(x + y + 1) + 1$.
Alternatively, we could leave it as $x^2 - (y + 1)^2 + 1$, which is also a valid simplified form.
Since the problem asks to "simplify," and factoring is a common way to simplify, the factored form is likely preferred.
So, the final simplified form is $(x - y - 1)(x + y + 1) + 1$.

Now, let's move on to the second expression, $3x^2 - 2x + 4$.
This is a quadratic expression in one variable, x.
To simplify it, we can try to factor it.
We look for two numbers that multiply to $3 \times 4 = 12$ and add up to $-2$.
The possible pairs of factors of 12 are: (1, 12), (2, 6), (3, 4), (-1, -12), (-2, -6), (-3, -4).
None of these pairs add up to -2.
Therefore, the quadratic expression $3x^2 - 2x + 4$ cannot be factored over the integers.
We can also check the discriminant to confirm if it has real roots.
The discriminant, D, is given by $b^2 - 4ac$, where a=3, b=-2, c=4.
$D = (-2)^2 - 4(3)(4) = 4 - 48 = -44$.
Since the discriminant is negative, the quadratic has no real roots and cannot be factored into linear factors with real coefficients.
Therefore, the expression $3x^2 - 2x + 4$ is already in its simplest form.

In conclusion, the solutions are:
1. $x^2 - y^2 - 2y = (x - y - 1)(x + y + 1) + 1$
2. $3x^2 - 2x + 4$ is already simplified.