Solve the equations:
Q. `(x-7)(x-3)(x+1)(x+5)=1680`.

Answer

To solve the equation \((x-7)(x-3)(x+1)(x+5)=1680\), we can follow these steps:

### Step 1: Rearranging the Equation
We start with the equation:
\[ (x-7)(x-3)(x+1)(x+5) = 1680 \]
We can rearrange it as:
\[ (x-7)(x+5)(x-3)(x+1) = 1680 \]

### Step 2: Pairwise Multiplication
Next, we can multiply the factors in pairs:
1. Multiply \((x-7)\) and \((x+5)\):
\[ (x-7)(x+5) = x^2 - 2x - 35 \]
2. Multiply \((x-3)\) and \((x+1)\):
\[ (x-3)(x+1) = x^2 - 2x - 3 \]

### Step 3: Substitute and Simplify
Now we can substitute:
Let \(t = x^2 - 2x\). Then we can rewrite the equation as:
\[ (t - 35)(t - 3) = 1680 \]
Expanding this gives:
\[ t^2 - 38t + 105 = 1680 \]
Rearranging leads to:
\[ t^2 - 38t - 1575 = 0 \]

### Step 4: Factor the Quadratic Equation
Now we will factor the quadratic equation:
\[ t^2 - 38t - 1575 = 0 \]
We look for two numbers that multiply to \(-1575\) and add to \(-38\). The numbers are \(-63\) and \(25\). Thus, we can factor it as:
\[ (t - 63)(t + 25) = 0 \]

### Step 5: Solve for \(t\)
Setting each factor to zero gives:
1. \(t - 63 = 0 \Rightarrow t = 63\)
2. \(t + 25 = 0 \Rightarrow t = -25\)

### Step 6: Substitute Back for \(x\)
Recall that \(t = x^2 - 2x\). We will substitute back to find \(x\):
1. For \(t = 63\):
\[ x^2 - 2x - 63 = 0 \]
Factoring gives:
\[ (x - 9)(x + 7) = 0 \Rightarrow x = 9 \text{ or } x = -7 \]

2. For \(t = -25\):
\[ x^2 - 2x + 25 = 0 \]
The discriminant \(D = (-2)^2 - 4(1)(25) = 4 - 100 = -96\) is negative, indicating complex roots. Using the quadratic formula:
\[ x = \frac{-(-2) \pm \sqrt{-96}}{2(1)} = \frac{2 \pm 4i\sqrt{6}}{2} = 1 \pm 2i\sqrt{6} \]

### Final Solutions
The real solutions are:
\[ x = 9 \quad \text{and} \quad x = -7 \]
The complex solutions are:
\[ x = 1 + 2i\sqrt{6} \quad \text{and} \quad x = 1 - 2i\sqrt{6} \]