Partial Fractions - Summary of Forms

Partial Fractions

Partial Fractions

In solving algebra related problems and questions, we may sometimes deal with rational functions. A rational function is basically an algebraic polynomial fraction, in which we have polynomials on both the numerator and denominator.

Partial fractions is one of the simplest and most effective method in solving algebra related problems regarding rational functions.

In partial fractions, we separate the polynomials in our rational function into simpler form of polynomials.

Some of the applications of partial fractions include the solving of integration problems with rational functions, the binomial expansion and also the arithmetic series and sequences.

There are a few basic forms we need to memorize in partial fractions:

1.\(\enspace\) The linear form:
\(\hspace{6em} {\small (ax + b)}\)
Example:

\({\large\frac{3x \ + \ 5}{(x \ + \ 1)(2x \ + \ 7)} \ \equiv \ \frac{A}{(x \ + \ 1)} + \frac{B}{(2x \ + \ 7)} }\)

2.\(\enspace\) The quadratic form of a linear factor:
\(\hspace{6em} {\small (cx \ + \ d)^{2}} \)
Example:

\(\frac{3x + 5}{(x + 1){(2x + 7)}^{2}} \equiv \frac{A}{(x + 1)} + {\small\boxed{\frac{B}{(2x + 7)} + \frac{C}{{(2x + 7)}^{2}}}} \)

3.\(\enspace\) The quadratic form that cannot be factorized:
\(\hspace{6em} {\small (c{x}^{2} \ + \ d) }\)
Example:

\({\large\frac{3x \ + \ 5}{(x \ + \ 1)(2{x}^{2} \ + \ 7)} \ \equiv \ \frac{A}{(x \ + \ 1)} + {\small\boxed{\frac{Bx \ + \ C}{(2{x}^{2} \ + \ 7)}}}}\)

After the polynomials in the denominator of the rational function is separated, make the denominators of the simpler terms to be the same. This is typically done by multiplying the denominators together .

To find each of the coefficients in the numerator (A, B or C), we can use substitution method or equating the coefficient method.

In the substitution method, we substitute a value of x that we freely choose in the left hand side numerator and the right hand side numerator and then find the coefficients one by one.

While in the equating the coefficient method, we expand the right hand side numerator and then compare each of the coefficients in the right hand side numerator with the left hand side numerator.

Both methods will be shown in the solution of the examples below. Give it a try and if you need any help, just look at the solution I have written. Cheers ! =) .
\(\\[1pt]\)


EXAMPLE:

\({\small 1.\enspace}\) Let \({\small f(x) \ = \ {\large \frac{7{x}^{2} \ – \ 15x \ + \ 8}{(1 \ – \ 2x){(2 \ – \ x)}^{2} }} }\)
\(\\[1pt]\)
Express \({\small f(x) }\) in partial fractions.
\(\\[1pt]\)

\(\\[1pt]\)
\({\small 2.\enspace}\) Let \({\small f(x) \ = \ {\large \frac{ x \ – \ 4{x}^{2} }{(3 \ – \ x)(2 \ + \ {x}^{2}) }} }\)
\(\\[1pt]\)
Express \({\small f(x) }\) in partial fractions.
\(\\[1pt]\)

\(\\[1pt]\)
\({\small 3.\enspace}\) Let \({\small f(x) \ = \ {\large \frac{ 5{x}^{2} \ + \ x \ + \ 27 }{(2x \ + \ 1)( {x}^{2} \ + \ 9 ) }} }\)
\(\\[1pt]\)
Express \({\small f(x) }\) in partial fractions.
\(\\[1pt]\)

\(\\[1pt]\)
\({\small 4.\enspace}\) Let \({\small f(x) \ = \ {\large \frac{ 10 x \ + \ 9 }{(2x \ + \ 1){( 2x \ + \ 3 )}^{2} }} }\)
\(\\[1pt]\)
Express \({\small f(x) }\) in partial fractions.
\(\\[1pt]\)

\(\\[1pt]\)
\({\small 5.\enspace}\) Let \({\small f(x) \ = \ {\large \frac{ 2x(5 \ – \ x) }{(3 \ + \ x){( 1 \ – \ x )}^{2} }} }\)
\(\\[1pt]\)
Express \({\small f(x) }\) in partial fractions.
\(\\[1pt]\)

\(\\[1pt]\)


PRACTICE MORE WITH THESE QUESTIONS BELOW!

\({\small 1.\enspace}\) Express \({\small {\large \frac{7{x}^{2} \ – \ 3x \ + \ 2}{x({x}^{2} \ + \ 1) }} }\) in partial fractions.

\({\small 2. \enspace}\) Let \({\small f(x) \ = \ {\large \frac{ 5{x}^{2} \ + \ x \ + \ 6 }{(3 \ – \ 2x)({x}^{2} \ + \ 4 )}} }\)
Express \({\small f(x) }\) in partial fractions.

\({\small 3. \enspace}\) Express \({\small {\large \frac{2 \ – \ x \ + \ 8{x}^{2}}{(1 \ – \ x)(1 \ + \ 2x)(2 \ + \ x) }} }\) in partial fractions.

\({\small 4. \enspace}\) Let \({\small f(x) \ = \ {\large \frac{ {x}^{2} \ + \ 3x \ + \ 3 }{(x \ + \ 1)(x \ + \ 3 )}} }\)
\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Express f(x) in partial fractions.
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Hence show that,
\(\hspace{3em} {\small \displaystyle \int_{0}^{3} f(x) \ \mathrm{d}x = 3 \ – \ \frac{1}{2} \ln 2.}\)

\({\small 5. \enspace}\) Let \({\small f(x) \ = \ {\large \frac{ {x}^{2} \ – \ 8x \ + \ 9 }{(1 \ – \ x){(2 \ – \ x )}^{2}}} }\)
Express \({\small f(x) }\) in partial fractions.

\({\small 6. \enspace}\) Let \({\small f(x) \ = \ {\large \frac{ 2{x}^{2} \ – \ 7x \ – \ 1 }{(x \ – \ 2)({x}^{2} \ + \ 3 )}} }\)
Express \({\small f(x) }\) in partial fractions.

\(\\[12pt]{\small 7. \enspace}\) Express \({\small {\large \frac{ x \ + \ 5 }{(x \ + \ 1)({x}^{2} \ + \ 3 )}} }\) in the form:
\(\hspace{2em} {\large \frac{A}{( x \ + \ 1)} + \frac{Bx \ + \ C}{ ( {x}^{2} \ + \ 3) } } \)

\({\small 8. \enspace}\) Express in partial fractions \({\small {\large \frac{{x}^{4}}{ {x}^{4} \ – \ 1 }} }\)

\({\small 9. \enspace}\) Express in partial fractions \({\small {\large \frac{{x}^{3} \ + \ x \ – \ 1}{ {x}^{2} \ + \ {x}^{4} }} }\)

\(\\[10pt]{\small 10.\enspace}\) Express in partial fractions
\(\hspace{2em}{\small {\large \frac{x \ + \ 5}{ {x}^{3} \ + \ 5{x}^{2} \ + \ 7x \ + \ 3 }} }\)


As always, if you have any particular questions to discuss, leave it in the comment section below. Cheers =) .

15 comments
  1. Hi Mr Will
    Solution for question no. 2
    5x^2 + x + 6 / (3 – 2x) (x^2 + 4)
    I got
    A = 3
    B = – 1
    C = – 2

    1. Yes, it’s correct!
      Great job, Anonymous =)

      Cheers,
      Mr Will

  2. Number 2

  3. Integrate
    No.1 1/x(x^n+1)dx
    No.2(x^2+1)*(x^2+2)/(x^2+3)*(x^2+4)

  4. May you please guide me more on the tips used to substitute for the value of x because it’s the most challenging part inorder to eliminate other coefficients . Thankyou.

  5. Solutions for the practice questions
    No.1
    7x^2-3x +2 /x(x^+1)
    I failed to get the value to substitute for x so as to eliminate A.
    No.2
    5x^2+6
    (3-2x) (x^2+4)
    Ans
    Solving values for B and C was a problem
    Mr Will am not at using computer ,therefore please some advise more especially when it comes at Mathematica.
    From Judith, Have a Good Night

    1. Hi Judith,
      For no 1, there are two factors in the denominator: the linear factor, x and the quadratic factor that can’t be factorized, \({\small {x}^{2} \ + \ 1}\).
      \(\\[1pt]\)
      Then we can use the appropriate forms,
      \(\\[1pt]\)
      \({\small {\large \frac{7{x}^{2} \ – \ 3x \ + \ 2}{x({x}^{2} \ + \ 1) } \equiv \frac{A}{x}+\frac{Bx \ + \ C}{ ({x}^{2} \ + \ 1) } } }\)
      \(\\[1pt]\)
      Next, multiply the denominators at the right hand side together.
      \(\\[1pt]\)
      \({\small {\large \frac{7{x}^{2} \ – \ 3x \ + \ 2}{x({x}^{2} \ + \ 1) } \equiv \frac{A({x}^{2} \ + \ 1) + (Bx \ + \ C)(x)}{x({x}^{2} \ + \ 1)} } }\)
      \(\\[1pt]\)
      Since the left hand side and the right hand side of the identity now have the same denominator, we can simply work on just the numerator.
      \(\\[1pt]\)
      \({\scriptsize 7{x}^{2} \ – \ 3x \ + \ 2 } \equiv {\small A({x}^{2} \ + \ 1) + (Bx \ + \ C)(x)} \)
      \(\\[1pt]\)
      We can use the substitution method,
      \(\\[10pt]\)Let \({\small x = 0}\),
      \(\\[5pt] {\small 7(0)^2 – 3(0) + 2 = A(0^2 + 1) }\)
      \(\\[5pt]\hspace{4.8 em} {\small 2 \ = \ A }\)
      \(\\[12pt]\hspace{4.9 em} {\small A \ = \ 2 }\)
      And we continue to substitute the x values to get B and C. Note that, you can pick any values of x arbitrarily. The idea is to get two simultaneous linear equations and solve for B and C.
      \(\\[1pt]\)
      \(\\[10pt]\)Let \({\small x = 1}\),
      \(\\[5pt] {\small 7(1)^2 – 3(1) + 2 = 2(1^2 + 1) \! + \! (B(1) + C)( 1 ) }\)
      \(\\[5pt]\hspace{4.8 em} {\small 6 \ = \ 4 \ + B + C }\)
      \(\\[12pt]\hspace{4.9 em} {\small 2 \ = \ B + C }\)
      \(\\[10pt]\)Let \({\small x = -1}\),
      \(\\[5pt] {\small 7(-1)^2 – 3(-1) + 2 = 2({-1}^{2} + 1) \! + \! (B(-1) + C)( -1 ) }\)
      \(\\[5pt]\hspace{4.8 em} {\small 12 \ = \ 4 \ + B \ – \ C }\)
      \(\\[12pt]\hspace{4.9 em} {\small 8 \ = \ B \ – \ C }\)
      By using substitution or elimination method, we can then find B and C,
      \(\\[10pt] \hspace{2 em} {\small B \ = \ 5}\)
      \(\\[10pt] \hspace{2 em} {\small C \ = \ -3}\)
      \(\\[10pt]\) Hence, f(x) in partial fractions is:
      \( f(x) \ = \ \frac{2}{x}+\frac{5x \ – \ 3}{ ({x}^{2} \ + \ 1) } \)
      \(\\[1pt]\)
      As for no 2, you can try it again Judith, just remember you can use the x value that make the factor (3 – 2x) to be zero to find the coefficient A and choose any values of x to find the coefficient B and C.
      Cheers,
      Will

  6. Hi Mr Will I am referring to the practice questions you gave after the presentation , so that I get to know what am practicing is write or not thankyou.

    1. Hi Judith,
      You can try the questions and maybe post your solutions here so I can take a look and see if there are any mistakes.
      Cheers

  7. please Mr Will solutions for the above question

    1. Hi Judith,
      which question are you referring about?

  8. 5x/(x+3)(x-2)

    1. \({\small {\large \frac{ 5x}{(x \ + \ 3)( x \ – \ 2 ) } \equiv \frac{A}{(x+3)}+\frac{B}{ (x \ – \ 2) } } }\)
      \(\\[1pt]\)
      \({\small {\large \frac{ 5x}{(x \ + \ 3)( x \ – \ 2 ) } \equiv \frac{A(x-2) \ + \ B(x+3)}{(x+3)(x \ – \ 2)} } }\)
      \(\\[1pt]\)
      \({\small 5x \equiv A(x-2) \ + \ B(x+3) }\)
      \(\\[1pt]\)
      \(\\[10pt]\)Let \({\small x = -3}\),
      \(\\[10pt]\hspace{1.2 em} {\small 5(-3) \ = \ A(-3 \ – \ 2) }\)
      \(\\[10pt]\hspace{2.2 em} {\small -15 \ = \ -5A }\)
      \(\hspace{3 em} {\small A \ = \ 3 }\)
      \(\\[1pt]\)
      \(\\[10pt]\)Let \({\small x = 2}\),
      \(\\[10pt]\hspace{1.2 em} {\small 5(2) \ = \ B(2 \ + \ 3) }\)
      \(\\[10pt]\hspace{2.4 em} {\small 10 \ = \ 5B }\)
      \(\hspace{3 em} {\small B \ = \ 2 }\)
      \(\\[1pt]\)
      \(\\[10pt]\) Hence,
      \( f(x) \ = \ \frac{3}{(x \ + \ 3)}+\frac{2}{ (x \ – \ 2) } \)

  9. If the roots of an equation x^2 -5x- 7=0 are (alpha symbol) and (beta symbol) find the equation whose roots are:
    (a).(alpha symbol)^2,(beta symbol)^2
    (b).(alpha symbol +1),(beta symbol +1)
    (c).(alpha symbol^2* beta symbol),(alpha symbol*beta symbol^2)

    1. The quadratic equation in its general form:
      \(\hspace{3em} {\small a{x}^{2} + bx + c \ = \ 0}\)
      with its roots: \({\small \alpha \ \textrm{and} \ \beta}\),
      has the formula for:
      \(\hspace{1.5em}\) the sum of roots, \(\enspace {\small \alpha \ + \ \beta \ = \ -\frac{b}{a}}\)
      \(\hspace{1.5em}\) the product of roots, \(\enspace {\small \alpha \ \times \ \beta \ = \ \frac{c}{a}}\)
      \(\\[1pt]\)
      And also, the new quadratic equation can be found if you know the sum of roots (s.r) and the product of roots (p.r),
      \(\enspace {\small {x}^{2} \ – \ \textrm{(s.r)}x \ + \ \textrm{(p.r)} \ = \ 0}\)
      \(\\[1pt]\)
      Now, back to the question,
      \(\hspace{3em} {\small {x}^{2} – 5x – 7 \ = \ 0}\)
      The sum of roots, \({\small \alpha \ + \ \beta \ = \ 5}\)
      The product of roots, \({\small \alpha \ \times \ \beta = -7}\)
      \(\\[1pt]\)
      \(\enspace\)(a). \( {\small {\alpha}^{2} \ + \ {\beta}^{2} \ = \ {(\alpha \ + \ \beta)}^2 \ – \ 2 \ \alpha \ \times \ \beta }\)
      \(\hspace{3em} \ {\small = \ 5^2 \ – \ 2 (-7)} \)
      \(\hspace{3em} \ {\small = \ 39 }\)
      \(\\[1pt]\)
      \(\hspace{1em} \ {\small {\alpha}^{2} \times \ {\beta}^{2} = {(\alpha \ \times \ \beta)}^{2} }\)
      \(\hspace{3em} \ {\small = \ (-7)^2 } \)
      \(\hspace{3em} \ {\small = \ 49 } \)
      \(\\[1pt]\)
      The new equation is \({\small x^2 \ – \ 39x \ + \ 49 = 0 }\)
      \(\\[1pt]\)
      \(\enspace\)(b). \( {\small (\alpha + 1) \ + \ (\beta + 1) \ = \ \alpha \ + \ \beta \ + \ 2 }\)
      \(\hspace{3em} \ {\small = \ 5 \ + \ 2 } \)
      \(\hspace{3em} \ {\small = \ 7 }\)
      \(\\[1pt]\)
      \( \ {\small (\alpha + 1) \ \times \ (\beta + 1) = \alpha \times \beta + \alpha + \beta + 1 }\)
      \(\hspace{3em} \ {\small = \ -7 \ + \ 5 \ + \ 1 } \)
      \(\hspace{3em} \ {\small = \ -1 } \)
      \(\\[1pt]\)
      The new equation is \({\small x^2 \ – \ 7x \ – \ 1 = 0 }\)
      \(\\[1pt]\)
      \(\enspace\)(c). \( {\small {\alpha}^{2}\beta \ + \ \alpha{\beta}^{2} \ = \ \alpha \beta (\alpha \ + \ \beta) }\)
      \(\hspace{3em} \ {\small = \ -7 (5)} \)
      \(\hspace{3em} \ {\small = \ -35 }\)
      \(\\[1pt]\)
      \(\hspace{1em} \ {\small {\alpha}^{2}\beta \ \times \ \alpha{\beta}^{2} = {(\alpha \ \times \ \beta)}^{3} }\)
      \(\hspace{3em} \ {\small = \ (-7)^3 } \)
      \(\hspace{3em} \ {\small = \ -243 } \)
      \(\\[1pt]\)
      The new equation is \({\small x^2 \ + \ 35x \ – \ 243 = 0 }\)

      Cheers

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