Question

resolve into partial fraction 2/(x-2)(x+1)​

Answer 1

Answer:

2/(x-2)(x+1)=A/x-2 +B/x+1

Step-by-step explanation:

$\frac{a}{x - 2} + \frac{b}{x + 1}$

Answer 2

Step-by-step explanation:

$\begin{array}{l} \displaystyle\rm \frac{2}{(x-2)(x+1)}=\frac{A}{x-2}+\frac{B}{x+1} \\ \\\displaystyle\rm\frac{2}{(x-2)(x+1)}=\frac{A(x+1)+B(x-2)}{(x-2)(x+1)}</p><p> \\ \\\rm2=A(x+1)+B(x−2) \\ \\\rm2=Ax+A+Bx−2B \\\\\rm 2=(A+B)x+A−2B \end{array}$

$\begin{aligned}&A+B=0\\&A-2B=2\end{aligned}$

• A=−B
• SubstituteA=−B into A−2B=2.
• -3B=2

• Solve for B in −3B=2.

$\rm \: B=-\dfrac{2}{3}$

Therefore,

$\begin{aligned}&A=\frac{2}{3}\\&B=-\frac{2}{3}\end{aligned}$

8 Substitute A,B into the original expression.

$\\ \rm\frac{2}{(x-2)(x+1)}=\frac{2}{3(x-2)}-\frac{2}{3(x+1)}$

Partial Fraction Decomposition completes.

$\\ \boxed{\red{ \rm\frac{2}{3(x-2)}-\frac{2}{3(x+1)}}}$