https://file.aiquickdraw.com/mathbot/file/1723038824996uanc450g.jpg

The expression given in the image is:

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

To express this in partial fractions, we will decompose the given fraction into a sum of simpler fractions. Since the denominator is composed of \((x - 2)\) and \((x - 1)^2\), we can write the partial fraction decomposition as:

\[
\frac{2x^2 - 4x + 1}{(x - 2)(x - 1)^2} = \frac{A}{x - 2} + \frac{B}{x - 1} + \frac{C}{(x - 1)^2}
\]

Here, \(A\), \(B\), and \(C\) are constants to be determined.

### Step 1: Combine the right side over a common denominator
\[
\frac{A}{x - 2} + \frac{B}{x - 1} + \frac{C}{(x - 1)^2} = \frac{A(x - 1)^2 + B(x - 2)(x - 1) + C(x - 2)}{(x - 2)(x - 1)^2}
\]

### Step 2: Equate the numerators
Since the denominators are the same, we can equate the numerators:

\[
2x^2 - 4x + 1 = A(x - 1)^2 + B(x - 2)(x - 1) + C(x - 2)
\]

### Step 3: Expand and collect like terms
Expand the right side:

\[
A(x^2 - 2x + 1) + B(x^2 - 3x + 2) + C(x - 2)
\]

Expanding:

\[
Ax^2 - 2Ax + A + Bx^2 - 3Bx + 2B + Cx - 2C
\]

Combine like terms:

\[
(A + B)x^2 + (-2A - 3B + C)x + (A + 2B - 2C)
\]

### Step 4: Match coefficients with the original numerator
Now, match the coefficients of corresponding powers of \(x\) from both sides:

1. Coefficient of \(x^2\): \(A + B = 2\)
2. Coefficient of \(x\): \(-2A - 3B + C = -4\)
3. Constant term: \(A + 2B - 2C = 1\)

### Step 5: Solve the system of equations
We now solve the system:

1. \(A + B = 2\)
2. \(-2A - 3B + C = -4\)
3. \(A + 2B - 2C = 1\)

Let's solve this system.

1. From the first equation: \(B = 2 - A\).
2. Substitute \(B = 2 - A\) into the second equation:
   \[
   -2A - 3(2 - A) + C = -4 \implies -2A - 6 + 3A + C = -4 \implies A + C = 2
   \]
3. Substitute \(B = 2 - A\) into the third equation:
   \[
   A + 2(2 - A) - 2C = 1 \implies A + 4 - 2A - 2C = 1 \implies -A - 2C = -3 \implies A + 2C = 3
   \]

Now we have two simpler equations:

- \(A + C = 2\)
- \(A + 2C = 3\)

Subtracting the first equation from the second gives:

\[
(A + 2C) - (A + C) = 3 - 2 \implies C = 1
\]

Substituting \(C = 1\) into \(A + C = 2\):

\[
A + 1 = 2 \implies A = 1
\]

Finally, substitute \(A = 1\) into \(B = 2 - A\):

\[
B = 2 - 1 = 1
\]

### Step 6: Write the final partial fraction decomposition
The partial fraction decomposition is:

\[
\frac{2x^2 - 4x + 1}{(x - 2)(x - 1)^2} = \frac{1}{x - 2} + \frac{1}{x - 1} + \frac{1}{(x - 1)^2}
\]

Would you like more details or have any questions?

Here are 8 related questions:

1. How do you solve for coefficients in a partial fraction decomposition when dealing with complex roots?
2. What is the general form for partial fractions when the denominator has repeated irreducible quadratic factors?
3. Can partial fractions be used to solve integrals involving rational functions? How?
4. What is the significance of partial fraction decomposition in Laplace Transforms?
5. How do you perform partial fraction decomposition when the degree of the numerator is equal to or greater than the denominator?
6. How would the partial fraction decomposition change if there were imaginary roots in the denominator?
7. Can you apply partial fractions to decompose a rational function with higher-order polynomial factors in the denominator?
8. How do partial fractions assist in solving differential equations?

**Tip:** When solving partial fraction decompositions, always check if the numerator's degree is less than the denominator's degree; otherwise, polynomial long division may be required first.

Learn how to perform partial fraction decomposition on the expression 2x^2 - 4x + 1 / ((x - 2)(x - 1)^2). Step-by-step solution included, covering polynomial algebra and advanced algebraic techniques. Suitable for advanced high school students.

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