Solve the equations:
`(x+1)(x+2)(x+3)(x+4)=120`

Answer

To solve the equation \((x+1)(x+2)(x+3)(x+4) = 120\), we can follow these steps:

### Step 1: Expand the left-hand side
We can group the terms and expand:
\[ (x+1)(x+4) = x^2 + 5x + 4 \]
\[ (x+2)(x+3) = x^2 + 5x + 6 \]
Now, we can rewrite the equation as:
\[ (x^2 + 5x + 4)(x^2 + 5x + 6) = 120 \]

### Step 2: Substitute \(T\)
Let \(T = x^2 + 5x\). Then the equation becomes:
\[ (T + 4)(T + 6) = 120 \]

### Step 3: Expand the equation
Expanding the left-hand side:
\[ T^2 + 10T + 24 = 120 \]

### Step 4: Rearrange the equation
Rearranging gives:
\[ T^2 + 10T + 24 - 120 = 0 \]
\[ T^2 + 10T - 96 = 0 \]

### Step 5: Factor the quadratic equation
Now, we need to factor the quadratic:
\[ T^2 + 10T - 96 = 0 \]
We can factor this as:
\[ (T - 6)(T + 16) = 0 \]

### Step 6: Solve for \(T\)
Setting each factor to zero gives:
\[ T - 6 = 0 \quad \Rightarrow \quad T = 6 \]
\[ T + 16 = 0 \quad \Rightarrow \quad T = -16 \]

### Step 7: Substitute back for \(x\)
Recall that \(T = x^2 + 5x\). We will solve for \(x\) in both cases.

**Case 1: \(T = 6\)**
\[ x^2 + 5x - 6 = 0 \]
Factoring gives:
\[ (x - 1)(x + 6) = 0 \]
Thus, \(x = 1\) or \(x = -6\).

**Case 2: \(T = -16\)**
\[ x^2 + 5x + 16 = 0 \]
Calculating the discriminant:
\[ D = b^2 - 4ac = 5^2 - 4 \cdot 1 \cdot 16 = 25 - 64 = -39 \]
Since the discriminant is negative, there are no real solutions for this case.

### Final Solutions
The real solutions to the original equation are:
\[ x = 1 \quad \text{and} \quad x = -6 \]