Partial Fraction with Cubic Equation in the Denominator :
Here we are going to see some practice questions on partial fractions with cubic equation in the denominator.
Question 1 :
Resolve the following rational expression into partial fractions.
Solution :
Let us decompose the denominator into linear factors.
(6x^{2} - x + 1)/(x^{3} + x^{2} + x + 1)
= A/(x + 1) + (Bx + C)/(x^{2} + 1)
(6x^{2} - x + 1) = A(x^{2} + 1) + (Bx + C)(x + 1)
Coefficients of x^{2}, x and constants
6 = A + B -----(1)
-1 = B + C -----(2)
1 = A + C -----(3)
(1) - (2) A + B - (B + C) = 6 + 1 A - C = 7 -----(4) |
(3) + (4) A + C + A - C = 1 + 7 2A = 8 A = 8/2 = 4 |
By applying the value of A in the 3rd equation
4 + C = 1
C = 1 - 4 = -3
C = -3
By applying the value of B in the 2nd equation
B + C = -1
B - 3 = -1
B = -1 + 3
B = 2
Question 2 :
Resolve the following rational expression into partial fractions.
Solution :
(x - 5)/(x^{2} + 2x - 3) = (x - 5)/(x + 3) (x - 1)
(x - 5)/(x + 3) (x - 1) = A/(x + 3) + B/(x - 1)
x - 5 = A(x - 1) + B(x + 3)
When x = 1 1 - 5 = B (1 + 3) -4 = 4B B = -1 |
When x = -3 -3 - 5 = A (-3 - 1) -8 = -4A A = 2 |
Question 3 :
Resolve the following rational expression into partial fractions.
(7 + x) / (1 + x)(1 + x^{2})
Solution :
(7 + x) / (1 + x)(1 + x^{2}) = A/(1 + x) + (Bx + C)/(1 + x^{2})
7 + x = A(1 + x^{2}) + (Bx + C) (1 + x)
Equating the coefficients of x2, x and constants.
0 = A + B -----(1)
1 = B + C -----(2)
7 = A + C -----(3)
(1) - (2) A + B - (B + C) = 0 - 1 A + B - B - C = - 1 A - C = -1 ----(4) |
(3) + (4) A + C + A - C = 7 - 1 2A = 6 A = 3 |
By applying the value of A in the first equation, we get
3 + B = 0
B = -3
By applying the value of B in the second equation, we get+
-3 + C = 1
C = 1 + 3
C = 4
(7 + x) / (1 + x)(1 + x^{2}) = 3/(1 + x) + (-3x + 4)/(1 + x^{2})
After having gone through the stuff given above, we hope that the students would have understood, how to resolve rational expressions into partial fractions.
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