$7x^3-86x^2\+346x-455$

$\=a_1{\left(x-4\right)}^3\+a_2\left(x-3\right){\left(x-4\right)}^2\+a_3\left(x-3\right)\left(x-4\right)\+a_4\left(x-3\right)$

$\=\left(a_1\+a_2\right)x^3\+\left(-12a_1-11a_2\+a_3\right)x^2\+\left(48a_1\+40a_2-7a_3\+a_4\right)x-64a_1-48a_2\+12a_3-3a_4$

$a_1\+a_2$

$\=7$

$-12a_1-11a_2\+a_3$

$\=-86$

$48a_1\+40a_2-7a_3\+a_4$

$\=346$

$-64a_1-48a_2\+12a_3-3a_4$

$\=-455$

Solution by Context Panel

$\frac{7x^3-86x^2\+346x-455}{x^4-15x^3\+84x^2-208 x\+192}$$\to$$\frac{2}{x-3}\+\frac{5}{x-4}-\frac{7}{{\left(x-4\right)}^2}\+\frac{1}{{\left(x-4\right)}^3}$

Solution by Task Template

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

Table 6.4.3(a)   Task template for a stepwise interactive partial-fraction decomposition

Interactive solution from first principles

$\frac{7x^3-86x^2\+346x-455}{x^4-15x^3\+84x^2-208 x\+192}$$\stackrel{\text{assign to a name}}{\to }$$f$

$\frac{a_1}{x-3}\+\frac{a_2}{x-4}\+\frac{a_3}{{\left(x-4\right)}^2}\+\frac{a_4}{{\left(x-4\right)}^3}$$\stackrel{\text{assign to a name}}{\to }$$g$

$f-g$

$\frac{7x^3-86x^2\+346x-455}{x^4-15x^3\+84x^2-208x\+192}-\frac{a_1}{x-3}-\frac{a_2}{x-4}-\frac{a_3}{{\left(x-4\right)}^2}-\frac{a_4}{{\left(x-4\right)}^3}$

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

$-\frac{x^3{a}_1\+x^3{a}_2-7x^3-12x^2{a}_1-11x^2{a}_2\+x^2{a}_3\+86x^2\+48x{a}_1\+40x{a}_2-7x{a}_3\+x{a}_4-346x-64a_1-48a_2\+12a_3-3a_4\+455}{\left(x-3\right){\left(x-4\right)}^3}$

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

$-x^3{a}_1-x^3{a}_2\+7x^3\+12x^2{a}_1\+11x^2{a}_2-x^2{a}_3-86x^2-48x{a}_1-40x{a}_2\+7x{a}_3-x{a}_4\+346x\+64a_1\+48a_2-12a_3\+3a_4-455$

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

$\left(-a_1-a_2\+7\right)x^3\+\left(12a_1\+11a_2-a_3-86\right)x^2\+\left(-48a_1-40a_2\+7a_3-a_4\+346\right)x\+64a_1\+48a_2-12a_3\+3a_4-455$

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

$64a_1\+48a_2-12a_3\+3a_4-455\,12a_1\+11a_2-a_3-86\,-48a_1-40a_2\+7a_3-a_4\+346\,-a_1-a_2\+7$

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

$\left\{a_1\=2\,a_2\=5\,a_3\=-7\,a_4\=1\right\}$

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

$S$

$\genfrac{}{}{0ex}{}{g}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{S}$ = $\frac{2}{x-3}\+\frac{5}{x-4}-\frac{7}{{\left(x-4\right)}^2}\+\frac{1}{{\left(x-4\right)}^3}$

Table 6.4.3(b)   Interactive solution from first principles

$f≔\frac{7x^3-86x^2\+346x-455}{x^4-15x^3\+84x^2-208 x\+192}\:$

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

$\frac{2}{x-3}\+\frac{5}{x-4}-\frac{7}{{\left(x-4\right)}^2}\+\frac{1}{{\left(x-4\right)}^3}$

Table 6.4.3(c)   Use of the convert command to obtain a partial-fraction decomposition

Coded solution from first principles

$q_1≔f\=g$

$\frac{7x^3-86x^2\+346x-455}{x^4-15x^3\+84x^2-208x\+192}\=\frac{a_1}{x-3}\+\frac{a_2}{x-4}\+\frac{a_3}{{\left(x-4\right)}^2}\+\frac{a_4}{{\left(x-4\right)}^3}$

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

$7x^3-86x^2\+346x-455\=x^3{a}_1\+x^3{a}_2-12x^2{a}_1-11x^2{a}_2\+x^2{a}_3\+48x{a}_1\+40x{a}_2-7x{a}_3\+x{a}_4-64a_1-48a_2\+12a_3-3a_4$

Form the equations determined by identifying the coefficients of powers of $x^k\,k\=0\,1\,2\,3$.

$$\begin{gathered} \mathbf{for} k \mathbf{from} 0 \mathbf{to} 3 \mathbf{do} \\ Q_k≔\mathrm{map}\left(\mathrm{coeff}\,q_2\,x\,k\right)\; \\ \mathbf{end} \mathbf{do} \end{gathered}$$

$-455\=-64a_1-48a_2\+12a_3-3a_4$

$q_3≔\mathrm{solve}\left(\left\{Q_0\,Q_1\,Q_2\,Q_3\right\}\right)$

$\left\{a_1\=2\,a_2\=5\,a_3\=-7\,a_4\=1\right\}$

$\mathrm{eval}\left(q_1\,q_3\right)$

$\frac{7x^3-86x^2\+346x-455}{x^4-15x^3\+84x^2-208x\+192}\=\frac{2}{x-3}\+\frac{5}{x-4}-\frac{7}{{\left(x-4\right)}^2}\+\frac{1}{{\left(x-4\right)}^3}$

Table 6.4.3(d)   Coded solution from first principles

$\mathrm{solve}\left(\mathrm{identity}\left(q_1\,x\right)\,\left\{a_1\,a_2\,a_3\,a_4\right\}\right)$

$\left\{a_1\=2\,a_2\=5\,a_3\=-7\,a_4\=1\right\}$