## anonymous 4 years ago Partial fraction decomposition (6x^3+8x)/(5x-7)^2

$\frac{6x ^{3}+8x}{(5x-7)^{2}}=\frac{A}{5x-7}+\frac{B}{(5x-7)^{2}}$ $6x ^{3}+8x=A(5x-7)+B$ $6x ^{3}+8x=5Ax-7A+B$ 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 $6(1)^{3}+8(1)=5A(1)-7A+\frac{3458}{125}$ $14=-2A+\frac{3458}{125}$ $\frac{854}{125}=A$ $\frac{6x ^{3}+8x}{(5x-7)^{2}}=\frac{854}{125(5x-7)}+\frac{3458}{125(5x-7)^{2}}$