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Answer
\( A = 3 \), \( B = -1 \), and \( C = 6 \).
Solution
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
\[
A = 3,\quad B = -1,\quad C = 6
\]
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Simplify this solution The Deep Dive
To solve the equation \( 3 x^{2}+5 x+C=A(x+1)^{2}+B(x+1)+4 \), we can start by expanding the right-hand side. This leads to \( A(x^2 + 2x + 1) + B(x + 1) + 4 = Ax^2 + (2A + B)x + (A + B + 4) \). Now, we can equate coefficients from both sides. For the \( x^2 \) term, we get \( 3 = A \). For the \( x \) term, we have \( 5 = 2A + B \). Substituting \( A = 3 \) gives us \( 5 = 6 + B \), thus \( B = -1 \). Finally, for the constant term, \( C = A + B + 4 \) becomes \( C = 3 - 1 + 4 = 6 \). In conclusion, the values are \( A = 3 \), \( B = -1 \), and \( C = 6 \).
