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Summary

Partial fractions

​​In a nutshell

An algebraic fraction can be written in the form P(x)Q(x)\dfrac{P(x)}{Q(x)}Q(x)P(x)​, where P(x)P(x)P(x) and Q(x)Q(x)Q(x) are polynomials. The order of a polynomial is given by the highest power of xxx. If the order of P(x)P(x)P(x) is less than the order of Q(x)Q(x)Q(x), then it is a proper fraction, and can be split up into two or more partial fractions. The denominators of the partial fractions are based on the linear factors of Q(x)Q(x)Q(x).​

​​​


Partial fractions

When splitting a algebraic fraction into partial fractions, there are two methods, which are:

  • Substitution
  • Equating coefficients


Substitution

procedure

1.
Factorise the denominator into linear factors.
2.
Set AAA and BBB as numerators of your partial fractions.
3.
Multiply up the fractions.​​
4.
In order to find the values of AAA and BBB, substitute x=root of the factorx=\rm root \ of \ the \ factorx=root of the factor.


Example 1

Split 7x+4x2−49\cfrac{7x+4}{x^2-49}x2−497x+4​ into partial fractions using the substitution method.


First, factorise the denominator into linear factors.

7x+4x2−49=7x+4(x+7)(x−7)\cfrac{7x+4}{x^2-49}=\cfrac{7x+4}{(x+7)(x-7)}x2−497x+4​=(x+7)(x−7)7x+4​​​


Now, set A\it AA and B\it BB as numerators of the partial fractions.

​7x+4(x+7)(x−7)=Ax+7+Bx−7\cfrac{7x+4}{(x+7)(x-7)} = \cfrac{A}{x+7}+\cfrac{B}{x-7}(x+7)(x−7)7x+4​=x+7A​+x−7B​​​


Multiply up the fractions to eliminate the denominators.

​7x+4(x+7)(x−7)=Ax+7+Bx−77x+4(x+7)(x−7)=A(x−7)(x+7)(x−7)+B(x+7)(x+7)(x−7)7x+4=A(x−7)+B(x+7)\begin{aligned}\dfrac{7x+4}{(x+7)(x-7)} &= \dfrac{A}{x+7}+\dfrac{B}{x-7}\\\dfrac{7x+4}{(x+7)(x-7)} &=\cfrac{A(x-7)}{(x+7)(x-7)} + \cfrac{B(x+7)}{(x+7)(x-7)} \\\\7x+4 & = A(x-7) + B(x+7)\end {aligned}(x+7)(x−7)7x+4​(x+7)(x−7)7x+4​7x+4​=x+7A​+x−7B​=(x+7)(x−7)A(x−7)​+(x+7)(x−7)B(x+7)​=A(x−7)+B(x+7)​​​

​​​

In order to find AAA, substitute x=−7x=-7x=−7, so BBB is eliminated from the equation.

​7(−7)+4=A(−7−7)+B(−7+7)−49+4=−14AA=4514‾\begin{aligned}{7(-7)+4} &={A(-7-7) + B(-7+7)} \\-49 +4 &= -14A \\ A&= \underline{\dfrac {45}{14}}\end{aligned}7(−7)+4−49+4A​=A(−7−7)+B(−7+7)=−14A=1445​​​​​

​​​

Repeat the process using x=7x=7x=7​ in order to find BBB.

​7×7+4=A(7−7)+B(7+7)53=14BB=5314‾\begin{aligned}7 \times 7 + 4 &= A ( 7-7) + B (7+7) \\ 53 &= 14B \\ B&= \underline{ \dfrac{53}{14}}\end{aligned}7×7+453B​=A(7−7)+B(7+7)=14B=1453​​​​​

​53=14B53= 14B53=14B

B=5314‾\underline{B = \cfrac{53}{14}}B=1453​​​​​

​

Therefore, 7x+4x2−49=4514(x+7)+5314(x−7)‾\underline{\dfrac{7x+4}{x^2-49}=\dfrac{45}{14(x+7)} + \dfrac{53}{14(x-7)}}x2−497x+4​=14(x+7)45​+14(x−7)53​​​​


Equating coefficients

procedure

1.
Factorise the denominator into linear factors.
2.
Set AAA and BBB as numerators of your partial fractions.
3.
Multiply up the fractions.
4.
Expand all brackets and equate the coefficients of xxx and the constant terms, to find the values of AAA and BBB.


Example 2

Split 7x+4x2−49\cfrac{7x+4}{x^2-49}x2−497x+4​ into partial fractions using the equating coefficients method. 


First, factorise the denominator into linear factors.

7x+4x2−49=7x+4(x+7)(x−7)\cfrac{7x+4}{x^2-49} = \cfrac{7x+4}{(x+7)(x-7)}x2−497x+4​=(x+7)(x−7)7x+4​​​


Now, set A\it AA and B\it BB as numerators of the partial fractons.

7x+4(x+7)(x−7)=Ax+7+Bx−7\cfrac{7x+4}{(x+7)(x-7)}=\cfrac{A}{x+7}+\cfrac{B}{x-7}(x+7)(x−7)7x+4​=x+7A​+x−7B​


Multiply up the fractions to eliminate the denominators.

7x+4(x+7)(x−7)=Ax+7+Bx−77x+4(x+7)(x−7)=A(x−7)(x+7)(x−7)+B(x+7)(x+7)(x−7)7x+4=A(x−7)+B(x+7)\begin{aligned}\dfrac{7x+4}{(x+7)(x-7)} &= \dfrac{A}{x+7}+\dfrac{B}{x-7}\\\dfrac{7x+4}{(x+7)(x-7)} &=\cfrac{A(x-7)}{(x+7)(x-7)} + \cfrac{B(x+7)}{(x+7)(x-7)} \\\\7x+4 & = A(x-7) + B(x+7)\end {aligned}(x+7)(x−7)7x+4​(x+7)(x−7)7x+4​7x+4​=x+7A​+x−7B​=(x+7)(x−7)A(x−7)​+(x+7)(x−7)B(x+7)​=A(x−7)+B(x+7)​​​​

​​

Next, expand all the brackets, to find the coefficients.

7x+4=Ax−7A+Bx+7B7x+4 = Ax - 7A + Bx + 7B7x+4=Ax−7A+Bx+7B​​

7x+4=x(A+B)+7(B−A)7x+4 = x(A+B)+7(B-A)7x+4=x(A+B)+7(B−A)​​


Equate coefficients of x.

7x=x(A+B)7x = x(A+B)7x=x(A+B)​​


And constant terms.

4=7(B−A)4 = 7 (B-A)4=7(B−A)​​

​​

And solve both equations simultaneously.

​{7x=x(A+B)4=7(B−A)\begin{cases}7x = x(A+B) \\4= 7 (B-A) \end{cases}{7x=x(A+B)4=7(B−A)​​​


Giving:

​A=4514‾B=5314‾\underline{A=\cfrac{45}{14}} \hspace{10mm}\underline{B=\cfrac{53}{14}}A=1445​​B=1453​​



Note: You will have to split your fractions into as many terms (A, B, CA, \ B, \ CA, B, C) as factors in the denominator you have.


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FAQs - Frequently Asked Questions

What is the equating coefficients method for solving partial fractions, and how does it work?

The equating coefficients method is a way of finding the linear factors of partial fractions. 1. Factorise the denominator into linear factors. 2. Set A and B as numerators of your partial fractions. 3. Multiply up the fractions. 4. Expand all brackets and equate the coefficients of x and the constant terms in order to find the values of A and B.

What is the substitution method for solving partial fractions, and how does it work?

The substitution method is a way of finding the linear factors of partial fractions. 1. Factorise the denominator into linear factors. 2. Set A and B as numerators of your partial fractions. 3. Multiply up the fractions. 4. Substitute x=root of the factor in order to find the values of A and B.

How many methods there are to solve partial fractions?

There are two methods: 1. Substitution 2. Equating coefficients

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