$5x^3-11x^2\+18x\+1$

$\=\left(b_1 x\+c_1\right) \left(x^2-3 x\+5\right)\+\left(b_2 x\+c_2\right) \left(x^2-2 x\+3\right)$

$\=x^3{b}_1\+x^3{b}_2-3x^2{b}_1-2x^2{b}_2\+x^2{c}_1\+x^2{c}_2\+5x{b}_1\+3x{b}_2-3x{c}_1-2x{c}_2\+5c_1\+3c_2$

$b_1\+b_2$

$\=5$

$-3b_1-2b_2\+c_1\+c_2$

$\=-11$

$5b_1\+3b_2-3c_1-2c_2$

$\=18$

$5c_1\+3c_2$

$\=1$

Solution by Context Panel

$\frac{5x^3-11x^2\+18x\+1}{x^4-5x^3\+14x^2-19 x\+15}$$\to$$\frac{3x\+2}{x^2-3x\+5}\+\frac{2x-1}{x^2-2x\+3}$

$f≔\frac{5x^3-11x^2\+18x\+1}{x^4-5x^3\+14x^2-19 x\+15}\:$

$\mathrm{convert}\left(f\,\mathrm{parfrac}\,x\right)$

$\frac{3x\+2}{x^2-3x\+5}\+\frac{2x-1}{x^2-2x\+3}$

Table 6.4.4(a)   Use of the convert command to obtain a partial-fraction decomposition

Solution by Task Template

Tools≻Tasks≻Browse: Algebra≻Partial Fractions≻Stepwise

Table 6.4.4(b)   Task template for an interactive stepwise partial-fraction decomposition

Interactive solution from first principles

$\frac{5x^3-11x^2\+18x\+1}{x^4-5x^3\+14x^2-19 x\+15}$$\stackrel{\text{assign to a name}}{\to }$$f$

$$$\frac{b_{1}x\+c_1}{x^2-2x\+3}\+\frac{b_{2}x\+c_2}{x^2-3x\+5}$$\stackrel{\text{assign to a name}}{\to }$$g$

$f-g$

$\frac{5x^3-11x^2\+18x\+1}{x^4-5x^3\+14x^2-19x\+15}-\frac{x{b}_1\+c_1}{x^2-2x\+3}-\frac{x{b}_2\+c_2}{x^2-3x\+5}$

$\stackrel{\text{simplify}}{\=}$

$-\frac{x^3{b}_1\+x^3{b}_2-5x^3-3x^2{b}_1-2x^2{b}_2\+x^2{c}_1\+x^2{c}_2\+11x^2\+5x{b}_1\+3x{b}_2-3x{c}_1-2x{c}_2-18x\+5c_1\+3c_2-1}{\left(x^2-2x\+3\right)\left(x^2-3x\+5\right)}$

$\stackrel{\text{numerator}}{\to }$

$-x^3{b}_1-x^3{b}_2\+5x^3\+3x^2{b}_1\+2x^2{b}_2-x^2{c}_1-x^2{c}_2-11x^2-5x{b}_1-3x{b}_2\+3x{c}_1\+2x{c}_2\+18x-5c_1-3c_2\+1$

$\stackrel{\text{collect w.r.t. x}}{\=}$

$\left(-b_1-b_2\+5\right)x^3\+\left(3b_1\+2b_2-c_1-c_2-11\right)x^2\+\left(-5b_1-3b_2\+3c_1\+2c_2\+18\right)x-5c_1-3c_2\+1$

$\stackrel{\text{coefficients in x}}{\to }$

$-5c_1-3c_2\+1\,3b_1\+2b_2-c_1-c_2-11\,-5b_1-3b_2\+3c_1\+2c_2\+18\,-b_1-b_2\+5$

$\stackrel{\text{solve}}{\to }$

$\left\{b_1\=2\,b_2\=3\,c_1\=-1\,c_2\=2\right\}$

$\stackrel{\text{assign to a name}}{\to }$

$S$

$\genfrac{}{}{0ex}{}{g}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{S}$ = $\frac{3x\+2}{x^2-3x\+5}\+\frac{2x-1}{x^2-2x\+3}$$$

Table 6.4.4(c)   Interactive solution from first principles

Coded solution from first principles

$q_1≔f\=g$

$\frac{5x^3-11x^2\+18x\+1}{x^4-5x^3\+14x^2-19x\+15}\=\frac{x{b}_1\+c_1}{x^2-2x\+3}\+\frac{x{b}_2\+c_2}{x^2-3x\+5}$

$q_2≔\mathrm{simplify}\left(q_1\cdot \mathrm{denom}\left(f\right)\right)$

$5x^3-11x^2\+18x\+1\=x^3{b}_1\+x^3{b}_2-3x^2{b}_1-2x^2{b}_2\+x^2{c}_1\+x^2{c}_2\+5x{b}_1\+3x{b}_2-3x{c}_1-2x{c}_2\+5c_1\+3c_2$

$q_3≔\mathrm{map}\left(\mathrm{coeff}\,q_2\,x\,0\right)$

$1\=5c_1\+3c_2$

$q_4≔\mathrm{map}\left(\mathrm{coeff}\,q_2\,x\,1\right)$

$18\=5b_1\+3b_2-3c_1-2c_2$

$q_5≔\mathrm{map}\left(\mathrm{coeff}\,q_2\,x\,2\right)$

$-11\=-3b_1-2b_2\+c_1\+c_2$

$q_6≔\mathrm{map}\left(\mathrm{coeff}\,q_2\,x\,3\right)$

$5\=b_1\+b_2$

$q_7≔\mathrm{solve}\left(\left\{q_3\,q_4\,q_5\,q_6\right\}\right)$

$\left\{b_1\=2\,b_2\=3\,c_1\=-1\,c_2\=2\right\}$

$\mathrm{eval}\left(q_1\,q_7\right)$

$\frac{5x^3-11x^2\+18x\+1}{x^4-5x^3\+14x^2-19x\+15}\=\frac{3x\+2}{x^2-3x\+5}\+\frac{2x-1}{x^2-2x\+3}$

Table 6.4.4(d)   Coded solution from first principles

$\mathrm{solve}\left(\mathrm{identity}\left(q_1\,x\right)\,\left\{b_1\,c_1\,b_2\,c_2\right\}\right)$

$\left\{b_1\=2\,b_2\=3\,c_1\=-1\,c_2\=2\right\}$