Solve the following equation: `(x-1) (x-2)(x-3)(x-4)=15`

Answer

To solve the equation \((x-1)(x-2)(x-3)(x-4) = 15\), we will follow these steps:

### Step 1: Expand the left-hand side
First, we can group and expand the factors on the left-hand side. We can multiply \((x-1)\) and \((x-4)\), and then multiply \((x-2)\) and \((x-3)\):

\[ (x-1)(x-4) = x^2 - 5x + 4 \]
\[ (x-2)(x-3) = x^2 - 5x + 6 \]

### Step 2: Set up the equation
Now we substitute these expansions back into the equation:

\[ (x^2 - 5x + 4)(x^2 - 5x + 6) = 15 \]

### Step 3: Let \(t = x^2 - 5x\)
To simplify the equation, we can let \(t = x^2 - 5x\). Thus, the equation becomes:

\[ (t + 4)(t + 6) = 15 \]

### Step 4: Expand the equation
Now we expand the left-hand side:

\[ t^2 + 10t + 24 = 15 \]

### Step 5: Rearrange the equation
Next, we rearrange the equation to set it to zero:

\[ t^2 + 10t + 24 - 15 = 0 \]
\[ t^2 + 10t + 9 = 0 \]

### Step 6: Factor the quadratic equation
Now we can factor the quadratic equation:

\[ (t + 9)(t + 1) = 0 \]

### Step 7: Solve for \(t\)
Setting each factor to zero gives us:

\[ t + 9 = 0 \quad \Rightarrow \quad t = -9 \]
\[ t + 1 = 0 \quad \Rightarrow \quad t = -1 \]

### Step 8: Substitute back for \(x\)
Now we substitute back for \(t\):

1. For \(t = -9\):
\[ x^2 - 5x = -9 \quad \Rightarrow \quad x^2 - 5x + 9 = 0 \]

Using the quadratic formula:
\[ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 9}}{2 \cdot 1} = \frac{5 \pm \sqrt{25 - 36}}{2} = \frac{5 \pm \sqrt{-11}}{2} \]
This gives us complex roots:
\[ x = \frac{5 \pm i\sqrt{11}}{2} \]

2. For \(t = -1\):
\[ x^2 - 5x = -1 \quad \Rightarrow \quad x^2 - 5x + 1 = 0 \]

Again using the quadratic formula:
\[ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{5 \pm \sqrt{25 - 4}}{2} = \frac{5 \pm \sqrt{21}}{2} \]

### Final Solutions
Thus, the solutions to the original equation are:

1. \( x = \frac{5 + i\sqrt{11}}{2} \)
2. \( x = \frac{5 - i\sqrt{11}}{2} \)
3. \( x = \frac{5 + \sqrt{21}}{2} \)
4. \( x = \frac{5 - \sqrt{21}}{2} \)