Partial fractions questions with solutions are given here for practice. Partial fractions are the decomposition of rational polynomials. Sometimes, while dealing with rational polynomials, it is not that easy to perform calculations, but those calculations become easy we decompose the complex rational polynomial function into simpler proper rational polynomial expressions.
Thus, to convert a single rational polynomial into the sum of two or more irreducible simple rational expressions is called Partial fraction decomposition.
For example,
The rational functions on the R.H.S are the partial fraction decomposition of the rational function on the L.H.S.
Process of finding partial fractions:
I. If the given rational function is an improper fraction, that is, the degree of the denominator is less than the degree of the numerator. Divide the numerator by the denominator to convert it into a simpler form.
II. If the denominator is non-repeating linear polynomials, then the partial fraction will be of the form A/(ax + b), where the value of A has to be determined.
III. If the denominator is n times repeating linear polynomials, then the partial fraction will be of the form A1/(ax + b) + A2/(ax + b)2 + A3/(ax + b)3 + … + An/(ax + b)n, where the values of A1, A2, A3, …, An have to be determined.
IV. If the denominator is a non-repeating quadratic polynomial, its partial fraction will be of the form (Ax + B)/(ax2 + bx + c), that is numerator will be linear. The values of A and B are to be determined.
V. If the denominator is n times repeating quadratic polynomial, its partial fraction will be of the form (A1x + B1)/(ax2 + bx + c) + (A2x + B2)/(ax2 + bx + c)2 + … + (Anx + Bn)/(ax2 + bx + c)n, that is numerators will be linear. The values of Ai’s and Bi’s (i = 1, 2, …, n) are to be determined.
Partial Fraction Types:
|
S.No. |
Rational Function Type |
Partial Fraction Decomposition |
|
I |
Linear and distinct factors in the denominator
\(\begin{array}{l}\frac{x-a}{(x+a)(x+b)}\end{array} \)
|
\(\begin{array}{l}\frac{A}{(x+a)}+\frac{B}{(x+b)}\end{array} \)
|
|
II |
Linear repeated factors in the denominator
\(\begin{array}{l}\frac{x-a}{(x+a)^{n}(x+b)}\end{array} \)
|
\(\begin{array}{l}\frac{A_{1}}{x+a}+\frac{A_{2}}{(x+a)^{2}}+…+\frac{A_{n}}{(x+a)^{n}}+\frac{B}{(x+b)}\end{array} \)
|
|
III |
Quadratic non-repeating denominator
\(\begin{array}{l}\frac{x-a}{(x+a)(x^{2}+b)}\end{array} \)
|
\(\begin{array}{l}\frac{A}{x+a}+\frac{Bx+C}{x^{2}+b}\end{array} \)
|
|
IV |
Quadratic repeated denominator
\(\begin{array}{l}\frac{x-a}{(x^{2}+a)(x^{2}+b)^{2}}\end{array} \)
|
\(\begin{array}{l}\frac{Ax+B}{(x^{2}+a)}+\frac{Cx+D}{x^{2}+b}+\frac{Ex+F}{(x^{2}+b)^{2}}\end{array} \)
|
Learn more about how to find partial fraction decompositions.
Partial Fraction Questions with Solutions
Let us solve a few partial fraction questions. Check your answers with solution given here.
Question 1:
Resolve into partial fractions:
Solution:
The given rational function is improper so divide it once, we get
Now, 3x2 – 2x – 1 = (x – 1)(3x + 1)
Let,
Multiplying both sides by (x –1)(3x + 1), we get
8x – 4 = A(3x + 1) + B(x – 1)
Put x = 1 ⇒ x – 1 = 0, we get
A = 1.
Again put x = – ⅓ ⇒ 3x + 1 = 0, we get
B = 5.
Hence, the required partial fraction is
Question 2:
Resolve into partial fractions:
Solution:
Dividing once the given fraction we get,
Let,
Multiply both sides by (x – 2)(x + 3), we get
6x – 2 = A(x + 3) + B(x – 2)
Put x = 2, we get
A = 2
Again put x = – 3
B = 4
Therefore, the required partial fraction is
Question 3:
Resolve into partial fractions:
Solution:
First we factorise the denominator
x3 – 2x2 – 8x = x(x – 4)(x + 2)
Let,
Multiply both sides by x(x – 4)(x +2), we get
8x – 8 = A(x – 4)(x – 2) + Bx(x + 2) + Cx (x – 4)
Put x = 0, we get, A = 1
Put x = 4, we get, B = 1
Put x = –2, we get, C = –2
∴ the required partial fraction is
Questions 4:
Resolve into partial fractions:
Solution:
Let
Multiplying both sides by x4(x + 1), we get
1 = Ax3(x + 1) + Bx2(x + 1) + Cx(x + 1) + D(x + 1) + Ex4
Put x = 0, we get, D = 1
Put x = –1, we get, E = 1
Now comparing the coefficients of x4 on both sides, we get
A + 1 = 0 ⇒ A = – 1
Now comparing the coefficients of x3 on both sides, we get
B + (-1) = 0 ⇒ B = 1
Now comparing the coefficients of x2 on both sides, we get
C + 1 = 0 ⇒ C = –1
∴ the required partial fractions are
Question 5:
Resolve the following into partial fractions:
Solution:
Let
Multiply both sides by (x + 1)3 (x + 2), we get,
4x3 + 16x2 + 23x + 13 = A(x + 1)2 (x + 2) + B(x + 1) (x + 2) + C(x + 2) + D(x + 1)3
Put x = –2, we get, D = 1
Put x = –1, we get, C = 2
Comparing the coefficients of x3 on both sides, we get
A + (-1) = 4
A = 3
Comparing the coefficients of x2 on both sides, we get
4A + B + 3D = 16
⇒ 12 + B + 3 = 16
⇒ B = 1
∴ the required partial fraction is
Question 6:
Resolve the following into partial fractions:
Solution:
Let,
Multiply both sides by (x – 1)3, we het
x2 – x – 3 = A(x – 1)2 + B(x – 1) + C
Put x = 1, we get, C = – 3
Comparing the coefficients of x2 on both the sides we get,
A = 1
Comparing the coefficients of x on both the sides we get,
B – 2A = –1 ⇒ B – 2 = –1
⇒ B = 1
∴ the required partial fraction is
Question 7:
Resolve the following into partial fractions:
Solution:
First, we factorise the denominator,
x4 + x2 + 1 = (x2 – x + 1)(x2 + x + 1)
Then, let
Multiply both sides by x4 + x2 + 1, we get,
x2 + 1 = (Ax + B)(x2 + x + 1) + (Cx + D)(x2 – x + 1)
Comparing the coefficients of x3 : A + C = 0 …(i)
Comparing the coefficients of x2 : A + B – C + D = 1 …(ii)
Comparing the coefficients of x : A + B + C – D = 0 …(iii)
Comparing the constants : B + D = 1 …(iv)
Subtract (iv) from (ii)
A – C = 0 …(v)
Adding (i) and (v), we get A = 0, C = 0
Putting this value in (ii) and (iii) and adding them we get,
2B = 1 ⇒ B = ½
Consequently, D = ½
∴ the required partial fraction is
Question 8:
Resolve the following into partial fractions
Solution:
Let,
Multiply both sides by (x – 2)(x2 + 5), we get,
x2 + 3x – 1 = A(x2 + 5) + (Bx + C)(x – 2)
Put x = 2, we get, A = 1
Comparing the coefficients of x2: A + B = 1 ⇒ 1 + B = 1 ⇒ B = 0
Comparing the coefficients of x: –2B + C = 3 ⇒ C = 3
∴ the required partial fraction is
Question 9:
Resolve the following into partial fractions:
Solution:
Let,
Multiply both sides by (1 – x)(1 + x2)2, we get,
x2 = A(1 + x2)2 + (Bx + C)(1 – x)(1 + x2) + (Dx + E)(1 – x)
Put x = 1, we get, A = ¼
Comparing the coefficients of x4: A – B = 0 ….(i)
Comparing the coefficients of x3: B – C = 0 ….(ii)
Comparing the coefficients of x2: 2A – B + C – D = 1 …(iii)
Comparing the coefficients of x: A + C + E = 0 ….(v)
Comparing the constant terms: A + C + E = 0
From (i) B = ¼ and from (ii) C = ¼
From (iii) D = –½ and lastly E = –½
The required partial fraction is
Question 10:
Resolve the following into partial fractions:
Solution:
Let,
Multiply both sides by (x + 1)(x2 + 2)2 we get,
7 = A(x2 + 2)2 + (Bx + C)(x + 1)(x2 + 2) + (Dx + E)(x + 1)
Put x = –1, we get, A = 7/9
Comparing the coefficients of x4: A + B = 0 ⇒ B = –7/9
Comparing the coefficients of x3: B + C = 0 ⇒ C = 7/9
Comparing the coefficients of x2: 2A + 2B + C + D = 0 ⇒ D = –7/9
Comparing the coefficients of x: 2B + 2C + D + E = 0 ⇒ E = 7/9
The required partial fraction is
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Practice Questions on Partial Fractions
Resolve the following into partial fractions:
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