Question

Partial fraction decomposition

(6x^3+8x)/(5x-7)^2

Answer 1

\[\frac{6x ^{3}+8x}{(5x-7)^{2}}=\frac{A}{5x-7}+\frac{B}{(5x-7)^{2}}\]
\[6x ^{3}+8x=A(5x-7)+B\]
\[6x ^{3}+8x=5Ax-7A+B\]
A zero of the original denominator is 7/5 so let x = 7/5\[6(\frac{7}{5})^{3}+8(\frac{7}{5})=5A(\frac{7}{5})-7A+B\]
\[B=\frac{3458}{125}\]
Let x = 1 for convenience
\[6(1)^{3}+8(1)=5A(1)-7A+\frac{3458}{125}\]
\[14=-2A+\frac{3458}{125}\]
\[\frac{854}{125}=A\]
\[\frac{6x ^{3}+8x}{(5x-7)^{2}}=\frac{854}{125(5x-7)}+\frac{3458}{125(5x-7)^{2}}\]