Answer :
Sure, let's tackle each part of the question one by one and decompose the given fractions into partial fractions.
### Part (a)
We need to decompose the fraction [tex]\(\frac{9}{(x-2)(x-3)}\)[/tex].
1. Step 1: Set up the partial fraction form:
We know that [tex]\(\frac{9}{(x-2)(x-3)}\)[/tex] can be decomposed into the form:
[tex]\[ \frac{A}{x-2} + \frac{B}{x-3} \][/tex]
where [tex]\(A\)[/tex] and [tex]\(B\)[/tex] are constants we need to determine.
2. Step 2: Combine the fractions on the right-hand side:
[tex]\[ \frac{A}{x-2} + \frac{B}{x-3} = \frac{A(x-3) + B(x-2)}{(x-2)(x-3)} \][/tex]
3. Step 3: Equate the numerators:
Since the denominators are the same, we can equate the numerators:
[tex]\[ 9 = A(x-3) + B(x-2) \][/tex]
4. Step 4: Solve for [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
We need to find [tex]\(A\)[/tex] and [tex]\(B\)[/tex] such that the equation holds for all [tex]\(x\)[/tex].
- For [tex]\(x = 2\)[/tex]:
[tex]\[ 9 = A(2-3) + B(2-2) \implies 9 = -A \implies A = -9 \][/tex]
- For [tex]\(x = 3\)[/tex]:
[tex]\[ 9 = A(3-3) + B(3-2) \implies 9 = B \implies B = 9 \][/tex]
5. Step 5: Write the decomposed form:
Given the values of [tex]\(A\)[/tex] and [tex]\(B\)[/tex], we can write:
[tex]\[ \frac{9}{(x-2)(x-3)} = \frac{-9}{x-2} + \frac{9}{x-3} \][/tex]
So, the partial fraction decomposition of [tex]\(\frac{9}{(x-2)(x-3)}\)[/tex] is:
[tex]\[ \frac{9}{(x-2)(x-3)} = \frac{-9}{x-2} + \frac{9}{x-3} \][/tex]
### Part (b)
Now, let's decompose the fraction [tex]\(\frac{2x+1}{x^2-3x+2}\)[/tex].
1. Step 1: Factor the denominator:
The denominator [tex]\(x^2 - 3x + 2\)[/tex] can be factored as:
[tex]\[ x^2 - 3x + 2 = (x-1)(x-2) \][/tex]
2. Step 2: Set up the partial fraction form:
We know that [tex]\(\frac{2x+1}{(x-1)(x-2)}\)[/tex] can be decomposed into the form:
[tex]\[ \frac{A}{x-1} + \frac{B}{x-2} \][/tex]
where [tex]\(A\)[/tex] and [tex]\(B\)[/tex] are constants we need to determine.
3. Step 3: Combine the fractions on the right-hand side:
[tex]\[ \frac{A}{x-1} + \frac{B}{x-2} = \frac{A(x-2) + B(x-1)}{(x-1)(x-2)} \][/tex]
4. Step 4: Equate the numerators:
Since the denominators are the same, we can equate the numerators:
[tex]\[ 2x + 1 = A(x-2) + B(x-1) \][/tex]
5. Step 5: Solve for [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
We need to find [tex]\(A\)[/tex] and [tex]\(B\)[/tex] such that the equation holds for all [tex]\(x\)[/tex].
- For [tex]\(x = 1\)[/tex]:
[tex]\[ 2(1) + 1 = A(1-2) + B(1-1) \implies 3 = -A \implies A = -3 \][/tex]
- For [tex]\(x = 2\)[/tex]:
[tex]\[ 2(2) + 1 = A(2-2) + B(2-1) \implies 5 = B \implies B = 5 \][/tex]
6. Step 6: Write the decomposed form:
Given the values of [tex]\(A\)[/tex] and [tex]\(B\)[/tex], we can write:
[tex]\[ \frac{2x+1}{(x-1)(x-2)} = \frac{-3}{x-1} + \frac{5}{x-2} \][/tex]
So, the partial fraction decomposition of [tex]\(\frac{2x+1}{x^2-3x+2}\)[/tex] is:
[tex]\[ \frac{2x+1}{x^2-3x+2} = \frac{-3}{x-1} + \frac{5}{x-2} \][/tex]
### Part (a)
We need to decompose the fraction [tex]\(\frac{9}{(x-2)(x-3)}\)[/tex].
1. Step 1: Set up the partial fraction form:
We know that [tex]\(\frac{9}{(x-2)(x-3)}\)[/tex] can be decomposed into the form:
[tex]\[ \frac{A}{x-2} + \frac{B}{x-3} \][/tex]
where [tex]\(A\)[/tex] and [tex]\(B\)[/tex] are constants we need to determine.
2. Step 2: Combine the fractions on the right-hand side:
[tex]\[ \frac{A}{x-2} + \frac{B}{x-3} = \frac{A(x-3) + B(x-2)}{(x-2)(x-3)} \][/tex]
3. Step 3: Equate the numerators:
Since the denominators are the same, we can equate the numerators:
[tex]\[ 9 = A(x-3) + B(x-2) \][/tex]
4. Step 4: Solve for [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
We need to find [tex]\(A\)[/tex] and [tex]\(B\)[/tex] such that the equation holds for all [tex]\(x\)[/tex].
- For [tex]\(x = 2\)[/tex]:
[tex]\[ 9 = A(2-3) + B(2-2) \implies 9 = -A \implies A = -9 \][/tex]
- For [tex]\(x = 3\)[/tex]:
[tex]\[ 9 = A(3-3) + B(3-2) \implies 9 = B \implies B = 9 \][/tex]
5. Step 5: Write the decomposed form:
Given the values of [tex]\(A\)[/tex] and [tex]\(B\)[/tex], we can write:
[tex]\[ \frac{9}{(x-2)(x-3)} = \frac{-9}{x-2} + \frac{9}{x-3} \][/tex]
So, the partial fraction decomposition of [tex]\(\frac{9}{(x-2)(x-3)}\)[/tex] is:
[tex]\[ \frac{9}{(x-2)(x-3)} = \frac{-9}{x-2} + \frac{9}{x-3} \][/tex]
### Part (b)
Now, let's decompose the fraction [tex]\(\frac{2x+1}{x^2-3x+2}\)[/tex].
1. Step 1: Factor the denominator:
The denominator [tex]\(x^2 - 3x + 2\)[/tex] can be factored as:
[tex]\[ x^2 - 3x + 2 = (x-1)(x-2) \][/tex]
2. Step 2: Set up the partial fraction form:
We know that [tex]\(\frac{2x+1}{(x-1)(x-2)}\)[/tex] can be decomposed into the form:
[tex]\[ \frac{A}{x-1} + \frac{B}{x-2} \][/tex]
where [tex]\(A\)[/tex] and [tex]\(B\)[/tex] are constants we need to determine.
3. Step 3: Combine the fractions on the right-hand side:
[tex]\[ \frac{A}{x-1} + \frac{B}{x-2} = \frac{A(x-2) + B(x-1)}{(x-1)(x-2)} \][/tex]
4. Step 4: Equate the numerators:
Since the denominators are the same, we can equate the numerators:
[tex]\[ 2x + 1 = A(x-2) + B(x-1) \][/tex]
5. Step 5: Solve for [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
We need to find [tex]\(A\)[/tex] and [tex]\(B\)[/tex] such that the equation holds for all [tex]\(x\)[/tex].
- For [tex]\(x = 1\)[/tex]:
[tex]\[ 2(1) + 1 = A(1-2) + B(1-1) \implies 3 = -A \implies A = -3 \][/tex]
- For [tex]\(x = 2\)[/tex]:
[tex]\[ 2(2) + 1 = A(2-2) + B(2-1) \implies 5 = B \implies B = 5 \][/tex]
6. Step 6: Write the decomposed form:
Given the values of [tex]\(A\)[/tex] and [tex]\(B\)[/tex], we can write:
[tex]\[ \frac{2x+1}{(x-1)(x-2)} = \frac{-3}{x-1} + \frac{5}{x-2} \][/tex]
So, the partial fraction decomposition of [tex]\(\frac{2x+1}{x^2-3x+2}\)[/tex] is:
[tex]\[ \frac{2x+1}{x^2-3x+2} = \frac{-3}{x-1} + \frac{5}{x-2} \][/tex]
