\frac{1}{4-X}-\frac{3}{X^2-4\cdot \:X}=\frac{1+X}{8-2\cdot \:X}

The given equation is:

\[
\frac{1}{4 - X} - \frac{3}{X^2 - 4X} = \frac{1 + X}{8 - 2X}
\]

### Step 1: Simplify the terms.

#### Second term on the left-hand side:
The second term \(\frac{3}{X^2 - 4X}\) can be factored. We have:

\[
X^2 - 4X = X(X - 4)
\]

Thus, the term becomes:

\[
\frac{3}{X(X - 4)}
\]

#### Denominator on the right-hand side:
We can factor \(8 - 2X\) as:

\[
8 - 2X = 2(4 - X)
\]

Thus, the right-hand side becomes:

\[
\frac{1 + X}{2(4 - X)}
\]

### Step 2: Rewrite the equation.
Now we have the equation:

\[
\frac{1}{4 - X} - \frac{3}{X(X - 4)} = \frac{1 + X}{2(4 - X)}
\]

### Step 3: Find a common denominator for the left-hand side.
To combine the two fractions on the left-hand side, we need a common denominator. The first fraction has \(4 - X\) and the second fraction has \(X(X - 4)\). The common denominator is \(X(4 - X)\).

We rewrite the left-hand side as:

\[
\frac{1}{4 - X} - \frac{3}{X(X - 4)} = \frac{X - 3}{X(4 - X)}
\]

### Step 4: Set up the new equation.
Now, the equation becomes:

\[
\frac{X - 3}{X(4 - X)} = \frac{1 + X}{2(4 - X)}
\]

### Step 5: Cross-multiply to eliminate denominators.
Cross-multiplying both sides gives:

\[
2(4 - X)(X - 3) = X(4 - X)(1 + X)
\]

### Step 6: Expand both sides.
Expand both sides:

#### Left-hand side:
\[
2(4 - X)(X - 3) = 2[(4)(X - 3) - X(X - 3)] = 2[4X - 12 - X^2 + 3X] = 2[7X - X^2 - 12] = 14X - 2X^2 - 24
\]

#### Right-hand side:
\[
X(4 - X)(1 + X) = X[(4)(1 + X) - X(1 + X)] = X[4 + 4X - X - X^2] = X[4 + 3X - X^2] = 4X + 3X^2 - X^3
\]

### Step 7: Set up the final equation.
Now we have:

\[
14X - 2X^2 - 24 = 4X + 3X^2 - X^3
\]

### Step 8: Rearrange the terms.
Move all terms to one side:

\[
X^3 - 5X^2 - 10X + 24 = 0
\]

### Step 9: Solve the cubic equation.
We now have the cubic equation:

\[
X^3 - 5X^2 - 10X + 24 = 0
\]

To solve this, we can use methods like factoring, synthetic division, or the Rational Root Theorem.

Testing potential rational roots like \(X = 2\), we find:

\[
2^3 - 5(2^2) - 10(2) + 24 = 8 - 20 - 20 + 24 = -12 + 24 = 0
\]

Thus, \(X = 2\) is a root.

### Step 10: Factor the cubic equation.
Now that we know \(X = 2\) is a root, we can factor \(X^3 - 5X^2 - 10X + 24\) as \((X - 2)\) times a quadratic. Using synthetic division or polynomial division, we get:

\[
X^3 - 5X^2 - 10X + 24 = (X - 2)(X^2 - 3X - 12)
\]

### Step 11: Solve the quadratic.
Solve \(X^2 - 3X - 12 = 0\) using the quadratic formula:

\[
X = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-12)}}{2(1)} = \frac{3 \pm \sqrt{9 + 48}}{2} = \frac{3 \pm \sqrt{57}}{2}
\]

Thus, the solutions are:

\[
X = \frac{3 \pm \sqrt{57}}{2}
\]

### Final Solutions:
The solutions to the equation are:

\[
X = 2, \quad X = \frac{3 + \sqrt{57}}{2}, \quad X = \frac{3 - \sqrt{57}}{2}
\]

Let me know if you would like more details on any of the steps!

### Related Questions:
1. How do you solve cubic equations using synthetic division?
2. What is the Rational Root Theorem and how is it used in solving polynomial equations?
3. How do you factor cubic polynomials given one known root?
4. What are the possible methods to solve a quadratic equation?
5. How do you factor quadratic expressions when the discriminant is not a perfect square?

### Tip:
When solving polynomial equations, always test for possible rational roots first to simplify the factoring process.

\frac{1}{4 - X} - \frac{3}{X^2 - 4X} = \frac{1 + X}{8 - 2X}

Learn how to solve a rational equation involving cubic and quadratic terms with factoring and the Rational Root Theorem. This problem breaks down key steps like finding a common denominator, cross-multiplication, and solving the resulting cubic equation using synthetic division. Suitable for students in Grades 10-12.

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