Solve the equations:
Q. `x(x-1)(x+2)(x-3)+8=0`.

Answer

To solve the equation \( x(x-1)(x+2)(x-3) + 8 = 0 \), we will follow these steps:

### Step 1: Expand the equation
First, we need to expand the product \( x(x-1)(x+2)(x-3) \).

1. Multiply the first two factors:
\[ x(x-1) = x^2 - x \]

2. Multiply the last two factors:
\[ (x+2)(x-3) = x^2 - 3x + 2x - 6 = x^2 - x - 6 \]

3. Now, multiply the results from the first two steps:
\[ (x^2 - x)(x^2 - x - 6) = x^2(x^2 - x - 6) - x(x^2 - x - 6) \]
Expanding this gives:
\[ = x^4 - x^3 - 6x^2 - (x^3 - x^2 + 6x) = x^4 - 2x^3 - 5x^2 + 6x \]

4. Now, add 8 to the equation:
\[ x^4 - 2x^3 - 5x^2 + 6x + 8 = 0 \]

### Step 2: Factor the polynomial
Next, we will try to find rational roots using the Rational Root Theorem. We can test \( x = -1 \):

1. Substitute \( x = -1 \):
\[ (-1)^4 - 2(-1)^3 - 5(-1)^2 + 6(-1) + 8 = 1 + 2 - 5 - 6 + 8 = 0 \]
Since this equals zero, \( x + 1 \) is a factor.

### Step 3: Perform polynomial long division
Now, we will divide \( x^4 - 2x^3 - 5x^2 + 6x + 8 \) by \( x + 1 \):

1. Divide \( x^4 \) by \( x \) to get \( x^3 \).
2. Multiply \( x^3 \) by \( x + 1 \) to get \( x^4 + x^3 \).
3. Subtract this from the original polynomial:
\[ (-2x^3 - x^3) = -3x^3 \]
4. Bring down the next term:
\[ -3x^3 - 5x^2 + 6x + 8 \]
5. Repeat this process until you get the quotient:
\[ x^3 - 3x^2 - 2x + 8 \]

### Step 4: Find the roots of the cubic polynomial
Now we need to solve \( x^3 - 3x^2 - 2x + 8 = 0 \). We can test \( x = 2 \):

1. Substitute \( x = 2 \):
\[ 2^3 - 3(2^2) - 2(2) + 8 = 8 - 12 - 4 + 8 = 0 \]
So, \( x - 2 \) is another factor.

### Step 5: Perform polynomial long division again
Now divide \( x^3 - 3x^2 - 2x + 8 \) by \( x - 2 \):

1. Divide \( x^3 \) by \( x \) to get \( x^2 \).
2. Multiply \( x^2 \) by \( x - 2 \) to get \( x^3 - 2x^2 \).
3. Subtract:
\[ (-3x^2 + 2x^2) = -x^2 \]
4. Bring down the next term:
\[ -x^2 - 2x + 8 \]
5. Repeat this process until you get:
\[ x^2 - x - 4 \]

### Step 6: Solve the quadratic equation
Now we need to solve \( x^2 - x - 4 = 0 \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \( a = 1, b = -1, c = -4 \):
\[ x = \frac{1 \pm \sqrt{(-1)^2 - 4(1)(-4)}}{2(1)} = \frac{1 \pm \sqrt{1 + 16}}{2} = \frac{1 \pm \sqrt{17}}{2} \]

### Final solutions
The solutions to the original equation are:
1. \( x = -1 \)
2. \( x = 2 \)
3. \( x = \frac{1 + \sqrt{17}}{2} \)
4. \( x = \frac{1 - \sqrt{17}}{2} \)