How do you express $(3x)/(x^2+x-2)$ in partial fractions?

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Answer 1 · 2016-11-24T14:51:51

the partial fraction decomposition is $2/(x + 2) + 1/(x- 1)$

We can factor the denominator as $(x + 2)(x- 1)$ .

$A/(x + 2) + B/(x- 1) = (3x)/((x + 2)(x - 1))$

$(A(x- 1))/((x + 2)(x - 1)) + (B(x + 2))/((x + 2)(x - 1)) = (3x)/((x + 2)(x - 1))$

$Ax - A + Bx + 2B = 3x$

$(A + B)x + (2B - A) = 3x$

We can now write a system of equations.

${(A + B = 3), (2B - A = 0):}$

Solving:

$B = 3 - A$

$2(3 - A) - A = 0$

$6 - 2A - A = 0$

$-3A = -6$

$A = 2$

$A + B = 3$

$2 + B = 3$

$B = 1$

Hence, the partial fraction decomposition is $2/(x + 2) + 1/(x- 1)$ .

Hopefully this helps!

How do you express (3x)/(x^2+x-2)(3x)/(x^2+x-2) in partial fractions?