Math  /  Algebra

QuestionWrite the partial fraction decomposition 2x23x+13(x+2)(x22x+3)\frac{-2 x^{2}-3 x+13}{(x+2)\left(x^{2}-2 x+3\right)}

Studdy Solution

STEP 1

1. The given expression is a rational function where the degree of the numerator is less than the degree of the denominator.
2. The denominator can be factored into linear and/or irreducible quadratic factors.
3. The partial fraction decomposition involves expressing the rational function as a sum of simpler fractions.

STEP 2

1. Identify the form of the partial fraction decomposition.
2. Set up the equation for the partial fraction decomposition.
3. Solve for the unknown coefficients.

STEP 3

Identify the form of the partial fraction decomposition for the given rational function:
The denominator is (x+2)(x22x+3)(x+2)(x^2 - 2x + 3). The term x+2x+2 is a linear factor, and x22x+3x^2 - 2x + 3 is an irreducible quadratic factor. Therefore, the partial fraction decomposition will have the form:
2x23x+13(x+2)(x22x+3)=Ax+2+Bx+Cx22x+3\frac{-2x^2 - 3x + 13}{(x+2)(x^2 - 2x + 3)} = \frac{A}{x+2} + \frac{Bx + C}{x^2 - 2x + 3}

STEP 4

Set up the equation for the partial fraction decomposition by equating the original expression to the sum of the partial fractions:
2x23x+13=A(x22x+3)+(Bx+C)(x+2)-2x^2 - 3x + 13 = A(x^2 - 2x + 3) + (Bx + C)(x + 2)

STEP 5

Expand the right-hand side and collect like terms:
A(x22x+3)=Ax22Ax+3AA(x^2 - 2x + 3) = Ax^2 - 2Ax + 3A (Bx+C)(x+2)=Bx2+2Bx+Cx+2C(Bx + C)(x + 2) = Bx^2 + 2Bx + Cx + 2C Combine these:
Ax22Ax+3A+Bx2+2Bx+Cx+2C=(A+B)x2+(2A+2B+C)x+(3A+2C)Ax^2 - 2Ax + 3A + Bx^2 + 2Bx + Cx + 2C = (A + B)x^2 + (-2A + 2B + C)x + (3A + 2C)

STEP 6

Equate the coefficients of the expanded expression to the coefficients of the original numerator:
A+B=2A + B = -2 2A+2B+C=3-2A + 2B + C = -3 3A+2C=133A + 2C = 13

STEP 7

Solve the system of equations for AA, BB, and CC:
From A+B=2A + B = -2, express BB in terms of AA:
B=2A B = -2 - A
Substitute B=2AB = -2 - A into the second equation:
2A+2(2A)+C=3-2A + 2(-2 - A) + C = -3 2A42A+C=3-2A - 4 - 2A + C = -3 4A+C=1-4A + C = 1
Substitute B=2AB = -2 - A into the third equation:
3A+2C=133A + 2C = 13
Now solve the system:
1. 4A+C=1 -4A + C = 1
2. 3A+2C=13 3A + 2C = 13

From equation 1: C=4A+1 C = 4A + 1
Substitute into equation 2:
3A+2(4A+1)=133A + 2(4A + 1) = 13 3A+8A+2=133A + 8A + 2 = 13 11A=1111A = 11 A=1A = 1
Substitute A=1A = 1 back to find CC:
C=4(1)+1=5C = 4(1) + 1 = 5
Substitute A=1A = 1 into B=2AB = -2 - A:
B=21=3B = -2 - 1 = -3

STEP 8

Write the partial fraction decomposition using the values of AA, BB, and CC:
2x23x+13(x+2)(x22x+3)=1x+2+3x+5x22x+3\frac{-2x^2 - 3x + 13}{(x+2)(x^2 - 2x + 3)} = \frac{1}{x+2} + \frac{-3x + 5}{x^2 - 2x + 3}

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