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 side of the equation
We start by multiplying the factors on the left side. First, we can group the terms:

\[ (x-1)(x-4) \quad \text{and} \quad (x-2)(x-3) \]

Calculating \((x-1)(x-4)\):

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

Now calculating \((x-2)(x-3)\):

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

### Step 2: Combine the results
Now we combine the two results:

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

Let \(t = x^2 - 5x\). Then we rewrite the equation as:

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

### Step 3: Expand and simplify
Expanding the left side:

\[ t^2 + 6t + 4t + 24 = 15 \]

This simplifies to:

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

\[ t^2 + 10t + 9 = 0 \]

### Step 4: Factor the quadratic equation
Now we factor the quadratic equation:

\[ t^2 + 10t + 9 = (t + 1)(t + 9) = 0 \]

Setting each factor to zero gives us:

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

### Step 5: Substitute back for \(t\)
Recall that \(t = x^2 - 5x\). We substitute back to find \(x\):

1. For \(t = -1\):

\[ x^2 - 5x + 1 = 0 \]

Using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{5 \pm \sqrt{25 - 4}}{2} = \frac{5 \pm \sqrt{21}}{2} \]

2. For \(t = -9\):

\[ x^2 - 5x + 9 = 0 \]

Using the quadratic formula again:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{5 \pm \sqrt{25 - 36}}{2} = \frac{5 \pm \sqrt{-11}}{2} \]

This gives us imaginary solutions:

\[ x = \frac{5 \pm i\sqrt{11}}{2} \]

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

\[ x = \frac{5 \pm \sqrt{21}}{2} \quad \text{(real solutions)} \]
\[ x = \frac{5 \pm i\sqrt{11}}{2} \quad \text{(imaginary solutions)} \]