A polynomial in x of degree three vanishes when x = 1 and `x =- 2` and has the values 4 and 28 when `x =- 1` and `x = 2` respectively, then f(1) is

Answer

To solve the problem step by step, we will find the polynomial \( f(x) \) of degree three that satisfies the given conditions.

### Step 1: Formulate the Polynomial
Since the polynomial \( f(x) \) is of degree three and vanishes at \( x = 1 \) and \( x = -2 \), we can express it as:
\[ f(x) = k(x - 1)(x + 2)(ax + b) \]
where \( k \) is a constant, and \( ax + b \) is a linear factor that we need to determine.

### Step 2: Use Given Values
We are given two conditions:
1. \( f(-1) = 4 \)
2. \( f(2) = 28 \)

Let's substitute \( x = -1 \) into the polynomial:
\[ f(-1) = k(-1 - 1)(-1 + 2)(-a + b) = k(-2)(1)(-a + b) = 2k(a - b) \]
Setting this equal to 4 gives us:
\[ 2k(a - b) = 4 \quad \text{(Equation 1)} \]

Now, substitute \( x = 2 \):
\[ f(2) = k(2 - 1)(2 + 2)(2a + b) = k(1)(4)(2a + b) = 4k(2a + b) \]
Setting this equal to 28 gives us:
\[ 4k(2a + b) = 28 \quad \text{(Equation 2)} \]

### Step 3: Simplify the Equations
From Equation 1:
\[ k(a - b) = 2 \quad \text{(1)} \]
From Equation 2:
\[ k(2a + b) = 7 \quad \text{(2)} \]

### Step 4: Solve the System of Equations
Now we can express \( k \) from both equations:
From (1):
\[ k = \frac{2}{a - b} \]
From (2):
\[ k = \frac{7}{2a + b} \]

Setting these two expressions for \( k \) equal gives:
\[ \frac{2}{a - b} = \frac{7}{2a + b} \]

Cross-multiplying results in:
\[ 2(2a + b) = 7(a - b) \]
Expanding this:
\[ 4a + 2b = 7a - 7b \]
Rearranging gives:
\[ 4a + 2b + 7b - 7a = 0 \implies -3a + 9b = 0 \implies 3a = 9b \implies a = 3b \]

### Step 5: Substitute Back
Now we can substitute \( a = 3b \) back into either Equation 1 or 2. Using Equation 1:
\[ k(3b - b) = 2 \implies k(2b) = 2 \implies k = \frac{1}{b} \]

Substituting \( k \) into Equation 2:
\[ \frac{1}{b}(2(3b) + b) = 7 \implies \frac{1}{b}(6b + b) = 7 \implies \frac{7b}{b} = 7 \implies 7 = 7 \]
This is always true, so we can choose \( b = 1 \) (for simplicity). Then:
\[ a = 3b = 3 \cdot 1 = 3 \]

### Step 6: Final Polynomial
Now substituting \( a \) and \( b \) back into the polynomial:
\[ f(x) = k(x - 1)(x + 2)(3x + 1) \]
Choosing \( k = 1 \):
\[ f(x) = (x - 1)(x + 2)(3x + 1) \]

### Step 7: Find \( f(1) \)
Now we need to find \( f(1) \):
\[ f(1) = (1 - 1)(1 + 2)(3 \cdot 1 + 1) = 0 \cdot 3 \cdot 4 = 0 \]

Thus, the value of \( f(1) \) is:
\[ \boxed{0} \]