Factorise the expression `f(x)=2x^(3)-7x^(2)-3x+18`,
Hence, find all possible values of x for which f(x)=0.

Answer

To factorize the expression \( f(x) = 2x^3 - 7x^2 - 3x + 18 \) and find all possible values of \( x \) for which \( f(x) = 0 \), we can follow these steps:

### Step 1: Rearranging the expression
We start with the polynomial:
\[ f(x) = 2x^3 - 7x^2 - 3x + 18 \]

### Step 2: Grouping terms
We can rearrange and group the terms in a way that makes it easier to factor:
\[ f(x) = 2x^3 - 4x^2 - 3x^2 + 6x - 6x + 18 \]
This does not change the expression but helps us to factor by grouping.

### Step 3: Factoring by grouping
Now, we can group the terms:
\[ = (2x^3 - 4x^2) + (-3x^2 + 6x) + (-6x + 18) \]
Factoring out common terms from each group:
\[ = 2x^2(x - 2) - 3(x^2 - 6) \]

### Step 4: Factoring further
Next, we can factor out \( x - 2 \):
\[ = (x - 2)(2x^2 - 3x - 9) \]

### Step 5: Factoring the quadratic
Now we need to factor the quadratic \( 2x^2 - 3x - 9 \). We look for two numbers that multiply to \( 2 \times -9 = -18 \) and add to \( -3 \). The numbers are \( 3 \) and \( -6 \):
\[ = 2x^2 + 3x - 6x - 9 \]
Grouping again:
\[ = (2x^2 + 3x) + (-6x - 9) \]
Factoring out common terms:
\[ = x(2x + 3) - 3(2x + 3) \]
Thus,
\[ = (x - 3)(2x + 3) \]

### Step 6: Final factorization
Putting it all together, we have:
\[ f(x) = (x - 2)(2x + 3)(x - 3) \]

### Step 7: Finding the roots
To find the values of \( x \) for which \( f(x) = 0 \), we set each factor to zero:
1. \( x - 2 = 0 \) → \( x = 2 \)
2. \( 2x + 3 = 0 \) → \( 2x = -3 \) → \( x = -\frac{3}{2} \)
3. \( x - 3 = 0 \) → \( x = 3 \)

### Step 8: Listing all possible values
The possible values of \( x \) for which \( f(x) = 0 \) are:
- \( x = 2 \)
- \( x = -\frac{3}{2} \)
- \( x = 3 \)

### Final Answer
The possible values of \( x \) for which \( f(x) = 0 \) are \( 2, -\frac{3}{2}, \) and \( 3 \).

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