X(x+2)(x+4)(x+6)÷x(x+2) divide the following by the method of factorsisation 

To simplify the expression \(\frac{X(x+2)(x+4)(x+6)}{x(x+2)}\) using factorization, follow these steps:

### Step 1: Factor the expression
First, observe that the numerator and the denominator share common factors:
- The numerator: \(X(x+2)(x+4)(x+6)\)
- The denominator: \(x(x+2)\)

### Step 2: Cancel common factors
Since \((x+2)\) is a common factor in both the numerator and denominator, and \(x\) is also a factor of the numerator:

\[
\frac{X(x+2)(x+4)(x+6)}{x(x+2)} = X \cdot \frac{(x+2)(x+4)(x+6)}{x(x+2)} = X \cdot \frac{(x+4)(x+6)}{x}
\]

### Step 3: Simplify the expression
Now, the expression simplifies to:

\[
X \cdot \frac{(x+4)(x+6)}{x} = X \cdot \frac{(x^2 + 10x + 24)}{x}
\]

Finally, distribute \(X\) over the simplified expression:

\[
X \cdot \left(\frac{x^2}{x} + \frac{10x}{x} + \frac{24}{x}\right) = X \cdot (x + 10 + \frac{24}{x})
\]

Thus, the simplified form of the given expression is:

\[
X \cdot \left(x + 10 + \frac{24}{x}\right)
\]

### Final Answer:
\[
\frac{X(x+2)(x+4)(x+6)}{x(x+2)} = X \cdot \left(x + 10 + \frac{24}{x}\right)
\]

Would you like further details or explanations on this problem?

Here are some related questions:
1. What is factorization, and why is it useful in simplifying expressions?
2. How would the process change if \(X\) was a more complex expression?
3. What if the expression had additional terms in the numerator or denominator?
4. How do you handle factorization when dealing with polynomials of higher degrees?
5. What would happen if \(x+2\) did not cancel out in both numerator and denominator?
6. Can you explain how to factor and simplify rational expressions?
7. What other methods besides factorization could simplify this expression?
8. How would you graph the simplified expression?

**Tip:** Always look for common factors first when simplifying expressions—it often makes the problem much easier to solve!

Learn how to simplify the rational expression X(x+2)(x+4)(x+6)/x(x+2) using factorization. This problem involves factoring and simplifying rational expressions, suitable for advanced high school students.

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