Partial Fraction decomposition calculator

SolutionHelp

5x-4x2-x-2 $(5x-4)/(x^2-x-2)$

x-3x3+2x2+x $(x-3)/(x^3+2x^2+x)$

x2+1x(x-1)3 $(x^2+1)/(x(x-1)^3)$

Example

1. Partial Fraction 5x-4x2-x-2 $(5x-4)/(x^2-x-2)$

Solution:
1. Factors the denominator

5x-4x2-x-2=5x-4(x+1)(x-2) $(5x-4)/(x^2-x-2)=(5x-4)/((x+1)(x-2))$

2. Partial Fraction for each factors

∴5x-4(x+1)(x-2)=Ax+1+Bx-2 $:. (5x-4)/((x+1)(x-2))=A/(x+1)+B/(x-2)$

3. Multiply through by the common denominator of

(x+1)(x-2) $(x+1)(x-2)$

∴5x-4=A(x-2)+B(x+1) $:. 5x-4=A(x-2)+B(x+1)$

∴5x-4=Ax-2A+Bx+B $:. 5x-4=Ax-2A+Bx+B$

4. Group the

x $x$ -terms and the constant terms

∴5x-4=(A+B)x+(-2A+B) $:. 5x-4=(A+B)x+(-2A+B)$

5. Coefficients of the two polynomials must be equal, so we get equations

A+B=5 $A+B=5$

-2A+B=-4 $-2A+B=-4$

Solution of equations using Elimination method

Total Equations are

2 $2$

A+B=5→(1) $A+B=5 -> (1)$

-2A+B=-4→(2) $-2A+B=-4 -> (2)$

Select the equations

(1) $(1)$ and

(2) $(2)$ , and eliminate the variable

B $B$ .

Now use back substitution method
From (3)

3A=9 $3A=9$

⇒3A=9 $=>3A=9$

⇒A=93=3 $=>A=(9)/(3)=3$

From (1)

A+B=5 $A+B=5$

⇒(3)+B=5 $=>(3)+B=5$

⇒B+3=5 $=>B+3=5$

⇒B=5-3=2 $=>B=5-3=2$

Solution using back substitution method.

A=3,B=2 $A = 3,B = 2$

After solving these equations, we get

A=3,B=2 $A=3,B=2$

Substitute these values in the original fractions

(5x-4)(x+1)(x-2)=3x+1+2x-2 $((5x-4))/((x+1)(x-2))=(3)/(x+1)+(2)/(x-2)$