Student[Basics]
FactorSteps
generate steps in factoring polynomials
Calling Sequence
Parameters
Description
Examples
Compatibility
FactorSteps( expr, variable )
FactorSteps( expr, implicitmultiply = true )
expr
-
string or expression
variable
(optional) variable to collect the terms by
implicitmultiply
(optional) truefalse
output = ...
(optional) option to control the return value
displaystyle = ...
(optional) option to control the layout of the steps
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.
withStudent:-Basics:
FactorStepsx3+6x2+12x+8
x3+6⋅x2+12⋅x+8▫1. Factor using Trial Evaluations◦Rewrite in standard formx3+6x2+12x+8◦The factors of the constant coefficient8are:C=1,2,4,8◦Trial evaluations ofxin±Cfindx=−2satisfies the equation, sox+2is a factorx3+6x2+12x+8x=−2|x3+6x2+12x+8x=−2=0◦Divide byx+2x+2z2PPx2P+4xPP+4)x21x31+6x21+12x1+8x3+2x2.4x2+12x4x2+8x.4x+84x+8.0◦Quotient times divisor from long divisionx2+4x+4⋅x+2•2. Examine term:x2+4x+4▫3. Factor using the AC Method◦Examine quadraticx2+4x+4◦Look at the coefficients,Ax2+Bx+CA=1,B=4,C=4◦Find factors of |AC| = |1⋅4| =41,2,4◦Find pairs of the above factors, which, when multiplied equal41⋅4,2⋅2◦Which pairs of ± these factors have asumof B =4? Found:2+2=4◦Split the middle term to use above pairx2+2x+2x+4◦Factorxout of the first groupx⋅x+2+2x+4◦Factor2out of the second groupx⋅x+2+2⋅x+2◦x+2is a common factorx⋅x+2+2⋅x+2◦Group common factorx+2⋅x+2This gives:x+22•4. This gives:x+23
FactorStepsa2−b2
a2−1⋅b2•1. This is a difference of squares, in the formA2−B2a2−b2•2. Apply difference of squares rule:A2−B2=A+BA−Ba+ba−b
FactorStepsx2−x−12
x2−1⋅x−12▫1. Factor using the AC Method◦Rewrite in standard formx2−x−12◦Look at the coefficients,Ax2+Bx+CA=1,B=−1,C=−12◦Find factors of |AC| = |1⋅−12| =121,2,3,4,6,12◦Find pairs of the above factors, which, when multiplied equal121⋅12,2⋅6,3⋅4◦Which pairs of ± these factors have adifferenceof B =−1? Found:3−4=−1◦Split the middle term to use above pairx2+3x−4x−12◦Factorxout of the first groupx⋅x+3+−4x−12◦Factor−4out of the second groupx⋅x+3+−4⋅x+3◦x+3is a common factorx⋅x+3−4⋅x+3◦Group common factorx−4⋅x+3This gives:x−4⋅x+3
FactorSteps2y25+113y5+33
25⋅y2+1135⋅y+33•1. Remove rationals and common factor15⋅2y2+113y+165•2. Examine term:2y2+113y+165▫3. Factor using the AC Method◦Examine quadratic2y2+113y+165◦Look at the coefficients,Ay2+By+CA=2,B=113,C=165◦Find factors of |AC| = |2⋅165| =3301,2,3,5,6,10,11,15,22,30,33,55,66,110,165,330◦Find pairs of the above factors, which, when multiplied equal3301⋅330,2⋅165,3⋅110,5⋅66,6⋅55,10⋅33,11⋅30,15⋅22◦Which pairs of ± these factors have asumof B =113? Found:3+110=113◦Split the middle term to use above pair2y2+3y+110y+165◦Factoryout of the first groupy⋅2y+3+110y+165◦Factor55out of the second groupy⋅2y+3+55⋅2y+3◦2y+3is a common factory⋅2y+3+55⋅2y+3◦Group common factory+55⋅2y+3This gives:y+55⋅2y+3•4. This gives:15⋅y+55⋅2y+3
The Student[Basics][FactorSteps] command was introduced in Maple 2021.
For more information on Maple 2021 changes, see Updates in Maple 2021.
See Also
Student:-Basics
Student:-Basics:-ExpandSteps
Student:-Basics:-LinearSolveSteps
Student:-Calculus1:-ShowSolution
Student:-Calculus1:-ShowSteps
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