Math Problem Statement

solve by trick full answer x^2 +4 / (x-2)(x-3)^2 ChatGPT said: ChatGPT Let's solve the expression

𝑥 2 + 4 ( 𝑥 − 2 ) ( 𝑥 − 3 ) 2 (x−2)(x−3) 2

x 2 +4 ​

using the partial fraction decomposition method, along with some substitutions to quickly find the values of constants.

Step 1: Set Up Partial Fractions The decomposition for the given expression will be:

𝑥 2 + 4 ( 𝑥 − 2 ) ( 𝑥 − 3 ) 2

𝐴 𝑥 − 2 + 𝐵 𝑥 − 3 + 𝐶 ( 𝑥 − 3 ) 2 (x−2)(x−3) 2

x 2 +4 ​

x−2 A ​ + x−3 B ​ + (x−3) 2

C ​

Step 2: Eliminate the Denominator Multiply both sides by the denominator ( 𝑥 − 2 ) ( 𝑥 − 3 ) 2 (x−2)(x−3) 2 :

𝑥 2 + 4

𝐴 ( 𝑥 − 3 ) 2 + 𝐵 ( 𝑥 − 2 ) ( 𝑥 − 3 ) + 𝐶 ( 𝑥 − 2 ) x 2 +4=A(x−3) 2 +B(x−2)(x−3)+C(x−2) Step 3: Expand the Right-Hand Side Expand 𝐴 ( 𝑥 − 3 ) 2 A(x−3) 2 :

𝐴 ( 𝑥 2 − 6 𝑥 + 9 )

𝐴 𝑥 2 − 6 𝐴 𝑥 + 9 𝐴 A(x 2 −6x+9)=Ax 2 −6Ax+9A Expand 𝐵 ( 𝑥 − 2 ) ( 𝑥 − 3 ) B(x−2)(x−3):

𝐵 ( 𝑥 2 − 5 𝑥 + 6 )

𝐵 𝑥 2 − 5 𝐵 𝑥 + 6 𝐵 B(x 2 −5x+6)=Bx 2 −5Bx+6B Expand 𝐶 ( 𝑥 − 2 ) C(x−2):

𝐶 𝑥 − 2 𝐶 Cx−2C Now combine these to form:

𝑥 2 + 4

( 𝐴 + 𝐵 ) 𝑥 2 + ( − 6 𝐴 − 5 𝐵 + 𝐶 ) 𝑥 + ( 9 𝐴 + 6 𝐵 − 2 𝐶 ) x 2 +4=(A+B)x 2 +(−6A−5B+C)x+(9A+6B−2C) Step 4: Set Up a System of Equations Now, we compare coefficients from both sides:

For 𝑥 2 x 2 :

𝐴 + 𝐵

1 (1) A+B=1(1) For 𝑥 x:

− 6 𝐴 − 5 𝐵 + 𝐶

0 (2) −6A−5B+C=0(2) For the constant term:

9 𝐴 + 6 𝐵 − 2 𝐶

4 (3) 9A+6B−2C=4(3) Step 5: Solve the Equations Using Substitutions:

Substituting 𝑥

2 x=2:

2 2 + 4 ( 2 − 2 ) ( 2 − 3 ) 2  is undefined, but we can get  𝐶 (2−2)(2−3) 2

2 2 +4 ​  is undefined, but we can get C Setting 𝑥

2 x=2 in the equation gives:

4

𝐴 ( 2 − 3 ) 2 + 𝐵 ( 2 − 2 ) ( 2 − 3 ) + 𝐶 ( 2 − 2 ) 4=A(2−3) 2 +B(2−2)(2−3)+C(2−2) This tells us 𝐴 ( − 1 ) 2

4 A(−1) 2 =4 thus 𝐴

4 A=4.

Substituting 𝑥

3 x=3:

3 2 + 4 ( 3 − 2 ) ( 3 − 3 ) 2  is undefined, but we can solve for  𝐵 (3−2)(3−3) 2

3 2 +4 ​  is undefined, but we can solve for B Setting 𝑥

3 x=3 gives:

9 + 4

𝐴 ( 0 ) + 𝐵 ( 3 − 2 ) ( 0 ) + 𝐶 ( 3 − 2 ) 9+4=A(0)+B(3−2)(0)+C(3−2) So:

13

𝐶    ⟹    𝐶

13 13=C⟹C=13 Now substitute 𝐴

4 A=4 and 𝐶

13 C=13 into equation (1):

4 + 𝐵

1    ⟹    𝐵

1 − 4

− 3 4+B=1⟹B=1−4=−3 Final Values of Constants 𝐴

4 A=4 𝐵

− 3 B=−3 𝐶

13 C=13 Step 6: Write the Partial Fraction Decomposition So, the partial fraction decomposition is:

𝑥 2 + 4 ( 𝑥 − 2 ) ( 𝑥 − 3 ) 2

4 𝑥 − 2 + − 3 𝑥 − 3 + 13 ( 𝑥 − 3 ) 2 (x−2)(x−3) 2

x 2 +4 ​

x−2 4 ​ + x−3 −3 ​ + (x−3) 2

13 ​

Final Answer 𝑥 2 + 4 ( 𝑥 − 2 ) ( 𝑥 − 3 ) 2

4 𝑥 − 2 − 3 𝑥 − 3 + 13 ( 𝑥 − 3 ) 2 (x−2)(x−3) 2

x 2 +4 ​

x−2 4 ​ − x−3 3 ​ + (x−3) 2

13 ​

This is the simplified form of the original expression using partial fraction decomposition!

Solution

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Math Problem Analysis

Mathematical Concepts

Partial Fraction Decomposition
Algebra
Rational Functions

Formulas

Partial fraction decomposition: A/(x - a) + B/(x - b) + C/(x - b)^2
Solving systems of linear equations
Quadratic expansion: (x - a)^2 = x^2 - 2ax + a^2

Theorems

Method of Partial Fractions
System of Linear Equations

Suitable Grade Level

Grades 10-12