Calculatrice Décomposition partielle des fractions

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 Exemple

 Problèmes résolus

 Problèmes difficiles

Here, we show you a step-by-step solved example of partial fraction decomposition. This solution was automatically generated by our smart calculator:

$\frac{1}{x^2+2x-3}$

Factor the trinomial $x^2+2x-3$ finding two numbers that multiply to form $-3$ and added form $2$

$\begin{matrix}\left(-1\right)\left(3\right)=-3\\ \left(-1\right)+\left(3\right)=2\end{matrix}$

Rewrite the polynomial as the product of two binomials consisting of the sum of the variable and the found values

$\frac{1}{\left(x-1\right)\left(x+3\right)}$

Rewrite the fraction $\frac{1}{\left(x-1\right)\left(x+3\right)}$ in $2$ simpler fractions using partial fraction decomposition

$\frac{1}{\left(x-1\right)\left(x+3\right)}=\frac{A}{x-1}+\frac{B}{x+3}$

Find the values for the unknown coefficients: $A, B$. The first step is to multiply both sides of the equation from the previous step by $\left(x-1\right)\left(x+3\right)$

$1=\left(x-1\right)\left(x+3\right)\left(\frac{A}{x-1}+\frac{B}{x+3}\right)$

Multiplying polynomials

$1=\frac{\left(x-1\right)\left(x+3\right)A}{x-1}+\frac{\left(x-1\right)\left(x+3\right)B}{x+3}$

Simplifying

$1=\left(x+3\right)A+\left(x-1\right)B$

Assigning values to $x$ we obtain the following system of equations

$\begin{matrix}1=4A&\:\:\:\:\:\:\:(x=1) \\ 1=2A-2B&\:\:\:\:\:\:\:(x=-1)\end{matrix}$

Proceed to solve the system of linear equations

$\begin{matrix}4A & + & 0B & =1 \\ 2A & - & 2B & =1\end{matrix}$

Rewrite as a coefficient matrix

$\left(\begin{matrix}4 & 0 & 1 \\ 2 & -2 & 1\end{matrix}\right)$

Reducing the original matrix to a identity matrix using Gaussian Elimination

$\left(\begin{matrix}1 & 0 & \frac{1}{4} \\ 0 & 1 & -\frac{1}{4}\end{matrix}\right)$

The fraction $\frac{1}{\left(x-1\right)\left(x+3\right)}$ in decomposed fractions equals

$\frac{1}{4\left(x-1\right)}+\frac{-1}{4\left(x+3\right)}$

 Réponse finale au problème