Answer :
To perform the partial fraction decomposition of the rational expression
[tex]\[ \frac{2x^2 + 3x - 6}{x^3 - 6x - 9} \][/tex]
we start by factoring the denominator.
First, let's factor [tex]\(x^3 - 6x - 9\)[/tex]. We look for a rational root using the Rational Root Theorem. The possible rational roots for [tex]\(x^3 - 6x - 9 = 0\)[/tex] are [tex]\(\pm 1, \pm 3, \pm 9\)[/tex].
Testing these values:
- For [tex]\(x = 1\)[/tex]:
[tex]\[ 1^3 - 6(1) - 9 = 1 - 6 - 9 = -14 \neq 0 \][/tex]
- For [tex]\(x = -1\)[/tex]:
[tex]\[ (-1)^3 - 6(-1) - 9 = -1 + 6 - 9 = -4 \neq 0 \][/tex]
- For [tex]\(x = 3\)[/tex]:
[tex]\[ 3^3 - 6(3) - 9 = 27 - 18 - 9 = 0 \][/tex]
So [tex]\(x = 3\)[/tex] is a root.
Therefore, we can divide [tex]\(x^3 - 6x - 9\)[/tex] by [tex]\(x - 3\)[/tex]:
Using synthetic division for [tex]\((x^3 - 6x - 9) \div (x - 3)\)[/tex]:
[tex]\[ \begin{array}{r|rrrr} 3 & 1 & 0 & -6 & -9 \\ & & 3 & 9 & 9 \\ \hline & 1 & 3 & 3 & 0 \\ \end{array} \][/tex]
This shows:
[tex]\[ x^3 - 6x - 9 = (x - 3)(x^2 + 3x + 3) \][/tex]
So now the expression is:
[tex]\[ \frac{2x^2 + 3x - 6}{(x - 3)(x^2 + 3x + 3)} \][/tex]
To decompose it into partial fractions, we assume:
[tex]\[ \frac{2x^2 + 3x - 6}{(x - 3)(x^2 + 3x + 3)} = \frac{A}{x - 3} + \frac{Bx + C}{x^2 + 3x + 3} \][/tex]
Multiplying both sides by [tex]\((x - 3)(x^2 + 3x + 3)\)[/tex] to clear the denominators, we get:
[tex]\[ 2x^2 + 3x - 6 = A(x^2 + 3x + 3) + (Bx + C)(x - 3) \][/tex]
Expanding and combining like terms:
[tex]\[ 2x^2 + 3x - 6 = A(x^2 + 3x + 3) + Bx^2 - 3Bx + Cx - 3C \][/tex]
[tex]\[= Ax^2 + 3Ax + 3A + Bx^2 - 3Bx + Cx - 3C\][/tex]
[tex]\[= (A + B)x^2 + (3A - 3B + C)x + (3A - 3C)\][/tex]
Equating coefficients of corresponding powers of [tex]\(x\)[/tex]:
1. For [tex]\(x^2\)[/tex]:
[tex]\[ A + B = 2 \][/tex]
2. For [tex]\(x^1\)[/tex]:
[tex]\[ 3A - 3B + C = 3 \][/tex]
3. For the constant term:
[tex]\[ 3A - 3C = -6 \][/tex]
Solving this system of equations:
From [tex]\(3A - 3C = -6 \rightarrow A - C = -2\)[/tex]
From [tex]\(A + B = 2\)[/tex]
[tex]\[ B = 2 - A \][/tex]
Substituting [tex]\(B\)[/tex] in [tex]\(3A - 3(2 - A) + C = 3 \rightarrow 3A - 6 + 3A + C = 3 \rightarrow 6A + C - 6 = 3 \rightarrow 6A + C = 9 \rightarrow C = 9 - 6A\)[/tex]
So now we have:
[tex]\[ A - C = -2 \rightarrow A - (9 - 6A) = -2 \rightarrow A - 9 + 6A = -2 \rightarrow 7A - 9 = -2 \rightarrow 7A = 7 \rightarrow A = 1 \][/tex]
For [tex]\(B\)[/tex]:
[tex]\[ B = 2 - A = 2 - 1 = 1 \][/tex]
For [tex]\(C\)[/tex]:
[tex]\[ C = 9 - 6A = 9 - 6(1) = 3 \][/tex]
Thus, the partial fractions decomposition is:
[tex]\[ \frac{2x^2 + 3x - 6}{(x - 3)(x^2 + 3x + 3)} = \frac{1}{x - 3} + \frac{x + 3}{x^2 + 3x + 3} \][/tex]
[tex]\[ \frac{2x^2 + 3x - 6}{x^3 - 6x - 9} \][/tex]
we start by factoring the denominator.
First, let's factor [tex]\(x^3 - 6x - 9\)[/tex]. We look for a rational root using the Rational Root Theorem. The possible rational roots for [tex]\(x^3 - 6x - 9 = 0\)[/tex] are [tex]\(\pm 1, \pm 3, \pm 9\)[/tex].
Testing these values:
- For [tex]\(x = 1\)[/tex]:
[tex]\[ 1^3 - 6(1) - 9 = 1 - 6 - 9 = -14 \neq 0 \][/tex]
- For [tex]\(x = -1\)[/tex]:
[tex]\[ (-1)^3 - 6(-1) - 9 = -1 + 6 - 9 = -4 \neq 0 \][/tex]
- For [tex]\(x = 3\)[/tex]:
[tex]\[ 3^3 - 6(3) - 9 = 27 - 18 - 9 = 0 \][/tex]
So [tex]\(x = 3\)[/tex] is a root.
Therefore, we can divide [tex]\(x^3 - 6x - 9\)[/tex] by [tex]\(x - 3\)[/tex]:
Using synthetic division for [tex]\((x^3 - 6x - 9) \div (x - 3)\)[/tex]:
[tex]\[ \begin{array}{r|rrrr} 3 & 1 & 0 & -6 & -9 \\ & & 3 & 9 & 9 \\ \hline & 1 & 3 & 3 & 0 \\ \end{array} \][/tex]
This shows:
[tex]\[ x^3 - 6x - 9 = (x - 3)(x^2 + 3x + 3) \][/tex]
So now the expression is:
[tex]\[ \frac{2x^2 + 3x - 6}{(x - 3)(x^2 + 3x + 3)} \][/tex]
To decompose it into partial fractions, we assume:
[tex]\[ \frac{2x^2 + 3x - 6}{(x - 3)(x^2 + 3x + 3)} = \frac{A}{x - 3} + \frac{Bx + C}{x^2 + 3x + 3} \][/tex]
Multiplying both sides by [tex]\((x - 3)(x^2 + 3x + 3)\)[/tex] to clear the denominators, we get:
[tex]\[ 2x^2 + 3x - 6 = A(x^2 + 3x + 3) + (Bx + C)(x - 3) \][/tex]
Expanding and combining like terms:
[tex]\[ 2x^2 + 3x - 6 = A(x^2 + 3x + 3) + Bx^2 - 3Bx + Cx - 3C \][/tex]
[tex]\[= Ax^2 + 3Ax + 3A + Bx^2 - 3Bx + Cx - 3C\][/tex]
[tex]\[= (A + B)x^2 + (3A - 3B + C)x + (3A - 3C)\][/tex]
Equating coefficients of corresponding powers of [tex]\(x\)[/tex]:
1. For [tex]\(x^2\)[/tex]:
[tex]\[ A + B = 2 \][/tex]
2. For [tex]\(x^1\)[/tex]:
[tex]\[ 3A - 3B + C = 3 \][/tex]
3. For the constant term:
[tex]\[ 3A - 3C = -6 \][/tex]
Solving this system of equations:
From [tex]\(3A - 3C = -6 \rightarrow A - C = -2\)[/tex]
From [tex]\(A + B = 2\)[/tex]
[tex]\[ B = 2 - A \][/tex]
Substituting [tex]\(B\)[/tex] in [tex]\(3A - 3(2 - A) + C = 3 \rightarrow 3A - 6 + 3A + C = 3 \rightarrow 6A + C - 6 = 3 \rightarrow 6A + C = 9 \rightarrow C = 9 - 6A\)[/tex]
So now we have:
[tex]\[ A - C = -2 \rightarrow A - (9 - 6A) = -2 \rightarrow A - 9 + 6A = -2 \rightarrow 7A - 9 = -2 \rightarrow 7A = 7 \rightarrow A = 1 \][/tex]
For [tex]\(B\)[/tex]:
[tex]\[ B = 2 - A = 2 - 1 = 1 \][/tex]
For [tex]\(C\)[/tex]:
[tex]\[ C = 9 - 6A = 9 - 6(1) = 3 \][/tex]
Thus, the partial fractions decomposition is:
[tex]\[ \frac{2x^2 + 3x - 6}{(x - 3)(x^2 + 3x + 3)} = \frac{1}{x - 3} + \frac{x + 3}{x^2 + 3x + 3} \][/tex]
