3d) $$ 3 x^{2}+5 x+C=A(x+1)^{2}+B(x+1)+4 $$

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To solve the equation:

$$ 3x^{2} + 5x + C = A(x + 1)^{2} + B(x + 1) + 4 $$

we need to determine the values of $$ A $$, $$ B $$, and $$ C $$.

### Step 1: Expand the Right Side

First, expand the right-hand side of the equation:

$$
\begin{align*}
A(x + 1)^{2} + B(x + 1) + 4 &= A(x^2 + 2x + 1) + Bx + B + 4 \
&= A x^2 + 2A x + A + Bx + B + 4 \
&= A x^2 + (2A + B)x + (A + B + 4)
\end{align*}
$$

### Step 2: Equate Coefficients

Now, compare the coefficients of corresponding powers of $$ x $$ on both sides of the equation:

$$
\begin{cases}
\text{Coefficient of } x^2: & 3 = A \
\text{Coefficient of } x: & 5 = 2A + B \
\text{Constant term}: & C = A + B + 4
\end{cases}
$$

### Step 3: Solve for $$ A $$, $$ B $$, and $$ C $$

1. **Find $$ A $$:**
   $$
   3 = A \quad \Rightarrow \quad A = 3
   $$

2. **Find $$ B $$:**
   Substitute $$ A = 3 $$ into the second equation:
   $$
   5 = 2(3) + B \quad \Rightarrow \quad 5 = 6 + B \quad \Rightarrow \quad B = -1
   $$

3. **Find $$ C $$:**
   Substitute $$ A = 3 $$ and $$ B = -1 $$ into the third equation:
   $$
   C = 3 + (-1) + 4 = 6
   $$

### Final Solution

$$
A = 3,\quad B = -1,\quad C = 6
$$
$$ A = 3 $$, $$ B = -1 $$, and $$ C = 6 $$.

3d)

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