How do you decompose $\frac{- 4 y}{3 y^{2} - 4 y + 1}$ $(-4y)/(3y^2-4y+1)$ into partial fractions?

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Answer 1 · 2017-01-14T19:28:46

Please see the explanation for the process. The answer is:

$\frac{- 4 y}{3 y^{2} - 4 y + 1} = \frac{2}{3 y - 1} - \frac{2}{y - 1}$ $(-4y)/(3y^2 - 4y + 1) = 2/(3y - 1)-2/(y - 1)$

Factor the denominator $(y - 1) (3 y - 1)$ $(y - 1)(3y - 1)$

Write the equation of the partial fractions with the unknowns, A and B:

$\frac{- 4 y}{(y - 1) (3 y - 1)} = \frac{A}{y - 1} + \frac{B}{3 y - 1}$ $(-4y)/((y - 1)(3y - 1)) = A/(y - 1) + B/(3y - 1)$

Multiply both sides by the denominator:

$- 4 y = A (3 y - 1) + B (y - 1)$ $-4y = A(3y - 1) + B(y - 1)$

Make B disappear by setting y = 1:

$- 4 = A (3 - 1) + B (0)$ $-4 = A(3 - 1) + B(0)$

#A = -2

Substitute - 2 for A and set y = 0:

$- 4 (0) = - 2 (3 (0) - 1) + B (0 - 1)$ $-4(0) = -2(3(0) - 1) + B(0 - 1)$

$B = 2$ $B = 2$

Check:

$- \frac{2}{y - 1} + \frac{2}{3 y - 1}$ $-2/(y - 1) + 2/(3y - 1)$

$- \frac{2}{y - 1} \frac{3 y - 1}{3 y - 1} + \frac{2}{3 y - 1} \frac{y - 1}{y - 1}$ $-2/(y - 1)(3y - 1)/(3y - 1) + 2/(3y - 1)(y - 1)/(y - 1)$

$\frac{- 6 y + 2 + 2 y - 2}{(y - 1) (3 y - 1)}$ $(-6y + 2 + 2y - 2)/((y - 1)(3y - 1))$

$\frac{- 4 y}{(y - 1) (3 y - 1)}$ $(-4y)/((y - 1)(3y - 1))$

This checks.

How do you decompose −4y3y2−4y+1(-4y)/(3y^2-4y+1) into partial fractions?