Factorise : `x^(3)-23x^(2)+142x-120`

Answer

To factor the polynomial \( x^3 - 23x^2 + 142x - 120 \), we will use the trial and error method to find a root of the polynomial, and then we will factor it accordingly.

### Step 1: Find a root using trial and error
We will substitute small integer values into the polynomial to see if we can find a root (a value of \( x \) that makes the polynomial equal to zero).

1. **Evaluate \( f(1) \)**:
\[ f(1) = 1^3 - 23 \cdot 1^2 + 142 \cdot 1 - 120 \]
\[ = 1 - 23 + 142 - 120 = 0 \]
Since \( f(1) = 0 \), \( x = 1 \) is a root of the polynomial.

### Step 2: Write the factor corresponding to the root
Since \( x = 1 \) is a root, we can write one factor of the polynomial as \( (x - 1) \).

### Step 3: Perform polynomial long division
Now, we will divide the polynomial \( x^3 - 23x^2 + 142x - 120 \) by \( (x - 1) \) to find the other factors.

1. **Divide \( x^3 \) by \( x \)**:
\[ x^2 \]

2. **Multiply \( (x - 1) \) by \( x^2 \)**:
\[ x^3 - x^2 \]

3. **Subtract from the original polynomial**:
\[ (x^3 - 23x^2 + 142x - 120) - (x^3 - x^2) = -22x^2 + 142x - 120 \]

4. **Divide \( -22x^2 \) by \( x \)**:
\[ -22x \]

5. **Multiply \( (x - 1) \) by \( -22x \)**:
\[ -22x^2 + 22x \]

6. **Subtract**:
\[ (-22x^2 + 142x - 120) - (-22x^2 + 22x) = 120x - 120 \]

7. **Divide \( 120x \) by \( x \)**:
\[ 120 \]

8. **Multiply \( (x - 1) \) by \( 120 \)**:
\[ 120x - 120 \]

9. **Subtract**:
\[ (120x - 120) - (120x - 120) = 0 \]

### Step 4: Write the complete factorization
After performing the division, we find that:
\[ x^3 - 23x^2 + 142x - 120 = (x - 1)(x^2 - 22x + 120) \]

### Step 5: Factor the quadratic \( x^2 - 22x + 120 \)
Now we need to factor \( x^2 - 22x + 120 \).

1. **Find two numbers that multiply to \( 120 \) and add to \( -22 \)**:
The numbers are \( -10 \) and \( -12 \).

Thus, we can factor the quadratic as:
\[ x^2 - 22x + 120 = (x - 10)(x - 12) \]

### Final Factorization
Combining everything, we have:
\[ x^3 - 23x^2 + 142x - 120 = (x - 1)(x - 10)(x - 12) \]