The FactorSteps command accepts a polynomial and displays the steps required to factor the expression.

If expr is a string, then it is parsed into an expression using InertForm:-Parse so that no automatic simplifications are applied, and thus no steps are missed.  

The implicitmultiply option is only relevant when expr is a string.  This option is passed directly on to the InertForm:-Parse command and will cause things like 2x to be interpreted as 2*x, but also, xyz to be interpreted as x*y*z.

The output and displaystyle options are described in Student:-Basics:-OutputStepsRecord. The return value is controlled by the output option.  

This function is part of the Student:-Basics package.

$\mathrm{with}\left(\mathrm{Student}:-\mathrm{Basics}\right)\:$

$\mathrm{FactorSteps}\left(x^3+6x^2+12x+8\right)$

$\begin{array}{lll}& & x^3+6\cdot x^2+12\cdot x+8\\ \text{▫}& & \text{1. Factor using Trial Evaluations}\\ & \text{◦}& \text{Rewrite in standard form}\\ & & x^3+6x^2+12x+8\\ & \text{◦}& \text{The factors of the constant coefficient}8\text{are:}\\ & & C=\left\{1\,2\,4\,8\right\}\\ & \text{◦}& \text{Trial evaluations of}x\text{in}\text{±}C\text{find}x\text{=}\mathrm{-2}\text{satisfies the equation, so}x+2\text{is a factor}\\ & & \genfrac{}{}{0ex}{}{\left(x^3+6x^2+12x+8\right)}{\phantom{x=\mathrm{-2}}}|\genfrac{}{}{0ex}{}{\phantom{\left(x^3+6x^2+12x+8\right)}}{x=\mathrm{-2}}=0\\ & \text{◦}& \text{Divide by}x+2\\ & & \begin{array}{cc}\stackrel{\phantom{z^2}}{x+2}& \begin{array}{ccccc}& \phantom{\mathrm{PP}}{x}^2& \phantom{P}\+4x& \phantom{\mathrm{PP}}\+4& \\ \)\phantom{x^2}& \phantom{1}{x}^3& \phantom{1}\+6x^2& \phantom{1}\+12x& \phantom{1}\+8\\ & \multicolumn{2}{c}{\frac{x^3+2x^2}{\phantom{\.}}}& \\ & & \multicolumn{2}{c}{4x^2+12x}& & \\ & & \multicolumn{2}{c}{\frac{4x^2+8x}{\phantom{\.}}}& & \\ & & & \multicolumn{2}{c}{4x+8}& & & \\ & & & \multicolumn{2}{c}{\frac{4x+8}{\phantom{\.}}}& & & \\ & & & & 0\hfill & & & & \end{array}\end{array}\\ & \text{◦}& \text{Quotient times divisor from long division}\\ & & \left(x^2+4x+4\right)\cdot \left(x+2\right)\\ \text{•}& & \text{2. Examine term:}\\ & & x^2+4x+4\\ \text{▫}& & \text{3. Factor using the AC Method}\\ & \text{◦}& \text{Examine quadratic}\\ & & \left(x^2+4x+4\right)\\ & \text{◦}& \text{Look at the coefficients,}A{x}^2+Bx+C\\ & & \left[''A''=1\,''B''=4\,''C''=4\right]\\ & \text{◦}& \text{Find factors of |AC| = |}1\cdot 4\text{| =}4\\ & & \left\{1\,2\,4\right\}\\ & \text{◦}& \text{Find pairs of the above factors, which, when multiplied equal}4\\ & & \left\{1\cdot 4\,2\cdot 2\right\}\\ & \text{◦}& \text{Which pairs of ± these factors have a}\text{sum}\text{of B =}4\text{? Found:}\\ & & 2+2=4\\ & \text{◦}& \text{Split the middle term to use above pair}\\ & & x^2+\left(2x+2x\right)+4\\ & \text{◦}& \text{Factor}x\text{out of the first group}\\ & & \left(x\cdot \left(x+2\right)\right)+\left(2x+4\right)\\ & \text{◦}& \text{Factor}2\text{out of the second group}\\ & & x\cdot \left(x+2\right)+\left(2\cdot \left(x+2\right)\right)\\ & \text{◦}& x+2\text{is a common factor}\\ & & x\cdot \left(x+2\right)+2\cdot \left(x+2\right)\\ & \text{◦}& \text{Group common factor}\\ & & \left(x+2\right)\cdot \left(x+2\right)\\ & & \text{This gives:}\\ & & {\left(x+2\right)}^2\\ \text{•}& & \text{4. This gives:}\\ & & {\left(x+2\right)}^3\end{array}$

$\mathrm{FactorSteps}\left(a^2-b^2\right)$

$\begin{array}{lll}& & a^2-1\cdot b^2\\ \text{•}& & \text{1. This is a difference of squares, in the form}{A}^2-B^2\\ & & a^2-b^2\\ \text{•}& & \text{2. Apply difference of squares rule:}{A}^2-B^2=\left(A+B\right)\left(A-B\right)\\ & & \left(a+b\right)\left(a-b\right)\end{array}$

$\mathrm{FactorSteps}\left(x^2-x-12\right)$

$\begin{array}{lll}& & x^2-1\cdot x-12\\ \text{▫}& & \text{1. Factor using the AC Method}\\ & \text{◦}& \text{Rewrite in standard form}\\ & & x^2-x-12\\ & \text{◦}& \text{Look at the coefficients,}A{x}^2+Bx+C\\ & & \left[''A''=1\,''B''=\mathrm{-1}\,''C''=\mathrm{-12}\right]\\ & \text{◦}& \text{Find factors of |AC| = |}1\cdot \left(\mathrm{-12}\right)\text{| =}12\\ & & \left\{1\,2\,3\,4\,6\,12\right\}\\ & \text{◦}& \text{Find pairs of the above factors, which, when multiplied equal}12\\ & & \left\{1\cdot 12\,2\cdot 6\,3\cdot 4\right\}\\ & \text{◦}& \text{Which pairs of ± these factors have a}\text{difference}\text{of B =}\mathrm{-1}\text{? Found:}\\ & & 3-4=\mathrm{-1}\\ & \text{◦}& \text{Split the middle term to use above pair}\\ & & x^2+\left(3x-4x\right)-12\\ & \text{◦}& \text{Factor}x\text{out of the first group}\\ & & \left(x\cdot \left(x+3\right)\right)+\left(-4x-12\right)\\ & \text{◦}& \text{Factor}\mathrm{-4}\text{out of the second group}\\ & & x\cdot \left(x+3\right)+\left(\left(\mathrm{-4}\right)\cdot \left(x+3\right)\right)\\ & \text{◦}& x+3\text{is a common factor}\\ & & x\cdot \left(x+3\right)-4\cdot \left(x+3\right)\\ & \text{◦}& \text{Group common factor}\\ & & \left(x-4\right)\cdot \left(x+3\right)\\ & & \text{This gives:}\\ & & \left(x-4\right)\cdot \left(x+3\right)\end{array}$

$\mathrm{FactorSteps}\left(\frac{2y^2}{5}+\frac{113y}{5}+33\right)$

$\begin{array}{lll}& & \frac{2}{5}\cdot y^2+\frac{113}{5}\cdot y+33\\ \text{•}& & \text{1. Remove rationals and common factor}\\ & & \frac{1}{5}\cdot \left(2y^2+113y+165\right)\\ \text{•}& & \text{2. Examine term:}\\ & & 2y^2+113y+165\\ \text{▫}& & \text{3. Factor using the AC Method}\\ & \text{◦}& \text{Examine quadratic}\\ & & \left(2y^2+113y+165\right)\\ & \text{◦}& \text{Look at the coefficients,}A{y}^2+By+C\\ & & \left[''A''=2\,''B''=113\,''C''=165\right]\\ & \text{◦}& \text{Find factors of |AC| = |}2\cdot 165\text{| =}330\\ & & \left\{1\,2\,3\,5\,6\,10\,11\,15\,22\,30\,33\,55\,66\,110\,165\,330\right\}\\ & \text{◦}& \text{Find pairs of the above factors, which, when multiplied equal}330\\ & & \left\{1\cdot 330\,2\cdot 165\,3\cdot 110\,5\cdot 66\,6\cdot 55\,10\cdot 33\,11\cdot 30\,15\cdot 22\right\}\\ & \text{◦}& \text{Which pairs of ± these factors have a}\text{sum}\text{of B =}113\text{? Found:}\\ & & 3+110=113\\ & \text{◦}& \text{Split the middle term to use above pair}\\ & & 2y^2+\left(3y+110y\right)+165\\ & \text{◦}& \text{Factor}y\text{out of the first group}\\ & & \left(y\cdot \left(2y+3\right)\right)+\left(110y+165\right)\\ & \text{◦}& \text{Factor}55\text{out of the second group}\\ & & y\cdot \left(2y+3\right)+\left(55\cdot \left(2y+3\right)\right)\\ & \text{◦}& 2y+3\text{is a common factor}\\ & & y\cdot \left(2y+3\right)+55\cdot \left(2y+3\right)\\ & \text{◦}& \text{Group common factor}\\ & & \left(y+55\right)\cdot \left(2y+3\right)\\ & & \text{This gives:}\\ & & \left(y+55\right)\cdot \left(2y+3\right)\\ \text{•}& & \text{4. This gives:}\\ & & \frac{1}{5}\cdot \left(y+55\right)\cdot \left(2y+3\right)\end{array}$

The Student[Basics][FactorSteps] command was introduced in Maple 2021.

For more information on Maple 2021 changes, see Updates in Maple 2021.