Let f(x) = `2x^(2) + 5x + 1 `. If we write `f(x)` as `f(x) = a (x + 1) (x - 2) + b(x-2) (x - 1) + c(x - 1) (x +1)` for real numbers a , b, c then

Options

A. there are infinite number of choices for a , b, c
B. only one choice for a but infinite number of choices for b and c
C. exactly one choice for each of a, b, c
D. more than one but finite number of choices for a,b, c

Answer

To solve the problem, we need to express the quadratic function \( f(x) = 2x^2 + 5x + 1 \) in the form given:

\[ f(x) = a(x + 1)(x - 2) + b(x - 2)(x - 1) + c(x - 1)(x + 1) \]

### Step 1: Expand the terms

First, we will expand each of the terms in the expression.

1. **Expand \( a(x + 1)(x - 2) \)**:
\[ a(x + 1)(x - 2) = a(x^2 - 2x + x - 2) = a(x^2 - x - 2) = ax^2 - ax - 2a \]

2. **Expand \( b(x - 2)(x - 1) \)**:
\[ b(x - 2)(x - 1) = b(x^2 - x - 2x + 2) = b(x^2 - 3x + 2) = bx^2 - 3bx + 2b \]

3. **Expand \( c(x - 1)(x + 1) \)**:
\[ c(x - 1)(x + 1) = c(x^2 - 1) = cx^2 - c \]

### Step 2: Combine the expanded terms

Now, we combine all the expanded terms:

\[ f(x) = (ax^2 - ax - 2a) + (bx^2 - 3bx + 2b) + (cx^2 - c) \]

Combining like terms, we get:

\[ f(x) = (a + b + c)x^2 + (-a - 3b)x + (-2a + 2b - c) \]

### Step 3: Set coefficients equal to those in \( f(x) = 2x^2 + 5x + 1 \)

Now, we compare the coefficients from \( f(x) \) with those in \( 2x^2 + 5x + 1 \):

1. Coefficient of \( x^2 \):
\[ a + b + c = 2 \quad \text{(1)} \]

2. Coefficient of \( x \):
\[ -a - 3b = 5 \quad \text{(2)} \]

3. Constant term:
\[ -2a + 2b - c = 1 \quad \text{(3)} \]

### Step 4: Solve the system of equations

Now we have a system of three equations:

1. \( a + b + c = 2 \)
2. \( -a - 3b = 5 \)
3. \( -2a + 2b - c = 1 \)

We can express these equations in matrix form:

\[ \begin{bmatrix} 1 & 1 & 1 \\ -1 & -3 & 0 \\ -2 & 2 & -1 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 2 \\ 5 \\ 1 \end{bmatrix} \]

### Step 5: Calculate the determinant

To check if there is a unique solution, we calculate the determinant of the coefficient matrix:

\[ \text{Det} = \begin{vmatrix} 1 & 1 & 1 \\ -1 & -3 & 0 \\ -2 & 2 & -1 \end{vmatrix} \]

Calculating the determinant, we find:

\[ = 1 \cdot (-3 \cdot -1 - 0 \cdot 2) - 1 \cdot (-1 \cdot -1 - 0 \cdot -2) + 1 \cdot (-1 \cdot 2 - -3 \cdot -2) \]
\[ = 1 \cdot (3) - 1 \cdot (1) + 1 \cdot (-2 - 6) \]
\[ = 3 - 1 - 8 = -6 \]

Since the determinant is non-zero (\(-6 \neq 0\)), we conclude that there is a unique solution for \( a, b, c \).

### Final Answer

Thus, there is exactly one choice for each of \( a, b, c \).

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