// Numbas version: exam_results_page_options
{"name": "Dado que $(x -k)$ es un factor de $p(x) = x^3+bx^2+cx+d$.   Encuentra la factorizaci\u00f3n  de $p(x)$.", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Dado que $(x -k)$ es un factor de $p(x) = x^3+bx^2+cx+d$.   Encuentra la factorizaci\u00f3n  de $p(x)$.", "tags": [], "metadata": {"description": "", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "<p>El teorema del factor establece que si $ f (x) $ es un polinomio y $ f (p) = 0 $, entonces $ (x-p) $ es un factor de $ f (x) $.</p>", "advice": "<p>Para esta pregunta, se nos da que $(\\simplify{x+{z}})$ es un factor del polinomio</p>\n<p>$p(x) = \\simplify{x^3+({y}+{u}+{z})*x^2+({y}{u}+{z}{u}+{y}{z})*x+{y}{u}{z}},$</p>\n<p>y luego se nos pide que encontremos la factorizaci&oacute;n completa de $p(x) $.</p>\n<p>Sabemos que $ (\\simplify{x+{z}}) $ es un factor de $ p(x) $, por lo que podemos calcular los otros factores de $ p(x) $ a trav&eacute;s de una divisi&oacute;n larga.</p>\n<p>\\[<br/>\\begin{align}<br/>&amp;\\simplify{x^2+({u}+{y})x+{u}{y}}\\\\<br/>\\simplify{x+{z}} \\; &amp;\\overline{)\\simplify{x^3+({y}+{u}+{z})*x^2+({y}{u}+{z}{u}+{y}{z})*x+{y}{u}{z}}}\\\\<br/>&amp;\\;\\,<br/>\\simplify{x^3+{z}x^2}\\\\<br/>&amp;\\qquad\\quad<br/>\\overline{\\simplify[all,noLeadingMinus]{({u}+{y})x^2+({y}{u}+{z}{u}+{y}{z})*x+{y}{u}{z}}}\\\\<br/>&amp;\\qquad\\quad<br/>\\simplify[all,noLeadingMinus]{({u}+{y})x^2+({u}{z}+{z}{y})x}\\\\<br/>&amp;\\qquad\\quad\\quad\\quad\\quad<br/>\\overline{\\simplify[all,noLeadingMinus]{{y}{u}x+{y}{u}{z}}}\\\\<br/>&amp;\\qquad\\quad\\quad\\quad\\quad<br/>\\simplify[all,noLeadingMinus]{{y}{u}x+{y}{u}{z}}\\\\<br/>&amp;\\qquad\\qquad\\quad\\quad\\quad<br/>\\overline{0.}<br/>\\end{align}<br/>\\]</p>\n<p>Entonces podemos factorizar $\\simplify{x^2+({u}+{y})x+{u}{y}}$&nbsp;como</p>\n<p>\\[\\simplify{x^2+({u}+{y})x+{u}{y}} =(\\simplify{x+{y}})(\\simplify{x+{u}}).\\]</p>\n<p>Por lo tanto, la factorizaci&oacute;n completa de&nbsp; $p(x)$ es&nbsp;</p>\n<p>\\[<br/>\\begin{align}<br/>p(x) &amp;= \\simplify{x^3+({y}+{u}+{z})*x^2+({y}{u}+{z}{u}+{y}{z})*x+{y}{u}{z}},\\\\<br/>&amp;= (\\simplify{x+{y}})(\\simplify{x+{z}})(\\simplify{x+{u}}).<br/>\\end{align}<br/>\\]</p>", "rulesets": {}, "extensions": [], "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"u": {"name": "u", "group": "Ungrouped variables", "definition": "random(-2..3 except 0 except -y)", "description": "<p>Factor 2.</p>", "templateType": "anything", "can_override": false}, "y": {"name": "y", "group": "Ungrouped variables", "definition": "random(-2..3 except 0)", "description": "<p>Factor 3.</p>", "templateType": "anything", "can_override": false}, "z": {"name": "z", "group": "Ungrouped variables", "definition": "random(-3..3 except 0)", "description": "<p>Factor 1.</p>", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": "192"}, "ungrouped_variables": ["y", "u", "z"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "jme", "useCustomName": false, "customName": "", "marks": "2", "scripts": {"mark": {"script": "if(this.credit>0) {\n  // Parse the student's answer as a syntax tree\n  var studentTree = Numbas.jme.compile(this.studentAnswer,Numbas.jme.builtinScope);\n\n  // Create the pattern to match against \n  // we just want two sets of brackets, each containing two terms\n  // or one of the brackets might not have a constant term\n  // or for repeated roots, you might write (x+a)^2\n  var rule = Numbas.jme.compile('m_all(m_any(x,x+m_pm(m_number),(x+m_pm(m_number))^m_number))*m_nothing');\n\n  // Check the student's answer matches the pattern. \n  var m = Numbas.jme.display.matchTree(rule,studentTree,true);\n  // If not, take away marks\n  if(!m) {\n    this.multCredit(0.5,'Your answer is not fully factorised.');\n  }\n}", "order": "after"}}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "<p>Dado que $(\\simplify{x+{z}})$ es un factor de $p(x) = \\simplify{x^3+({y}+{u}+{z})*x^2+({y}*{u}+{z}*{u}+{y}*{z})*x+{y}*{u}*{z}}.$</p>\n<p>Encuentra la factorizaci&oacute;n completa de $p(x)$.</p>", "answer": "(x+{y})(x+{u})(x+{z})", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "x", "value": ""}]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always", "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Elliott Fletcher", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1591/"}, {"name": "Luis Hernandez", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2870/"}]}]}], "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Elliott Fletcher", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1591/"}, {"name": "Luis Hernandez", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2870/"}]}