Factoring Polynomials by Grouping

1. Group the first two terms and factor out $4x$:
\[ 4x^2 + 4x = 4x(x + 1) \]

2. Group the last two terms and factor out $3$:
\[ 3x + 3 = 3(x + 1) \]

3. Rewrite and factor out the common binomial $(x + 1)$:
\[ 4x(x + 1) + 3(x + 1) = (x + 1)(4x + 3) \]

1. Group terms with $x$: \( 2x^2 - 4x = 2x(x - 2) \)

2. Group terms with $y$: \( 3xy - 6y = 3y(x - 2) \)

3. Combine: \[ 2x(x - 2) + 3y(x - 2) = (x - 2)(2x + 3y) \]

Since there are only three terms, we split the middle term $4x$ into $3x + x$ to enable grouping:

\[ 3x^2 + 3x + x + 1 \]

Now group and factor:

\[ (3x^2 + 3x) + (x + 1) = 3x(x + 1) + 1(x + 1) \] \[ = (x + 1)(3x + 1) \]

\[ x(x + 3) - 2(x + 3) = (x + 3)(x - 2) \]

Graphical Check: The x-intercepts of the graph $y = x^2 + 3x - 2x - 6$ correspond to the roots $x = -3$ and $x = 2$.

Rewrite $x$ as $4x - 3x$:

\[ 4x^2 + 4x - 3x - 3 = 4x(x + 1) - 3(x + 1) = (x + 1)(4x - 3) \]

First, factor out the common $x$ from all terms:

\[ x(xy + 3 + xy^2 + 3y) \]

Now group inside the parentheses:

\[ x[xy(1 + y) + 3(1 + y)] = x(1 + y)(xy + 3) \]

Rewrite $-x$ as $-3x + 2x$ to create six terms:

\[ 3x^2 + 3xy - 3x + 2x + 2y - 2 \]

Group the first three and last three terms:

\[ 3x(x + y - 1) + 2(x + y - 1) = (x + y - 1)(3x + 2) \]