$\\simplify{(x+{a[0]})(x-{a[0]})}$ = [[0]]
\n\n", "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "gaps": [{"notallowed": {"message": "Ensure you don't use brackets in your answer.
", "showStrings": false, "strings": ["(", ")"], "partialCredit": 0}, "vsetrangepoints": 5, "expectedvariablenames": ["x"], "checkingaccuracy": 0.001, "vsetrange": [0, 1], "showpreview": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "scripts": {}, "answer": "x^2-{a[0]*a[0]}", "marks": 1, "checkvariablenames": true, "checkingtype": "absdiff", "type": "jme"}], "steps": [{"prompt": "Method 1 (the distributive law)
\nWe expand $\\simplify{(x+{a[0]})(x-{a[0]})}$ one bracket at a time.
\n| $\\simplify{(x+{a[0]})(x-{a[0]})}$ | \n$=$ | \n\n $\\simplify{x(x-{a[0]})+{a[0]}(x-{a[0]})}$ \n | \n\n (each term in one bracket times the entire other bracket) \n | \n
| \n | $=$ | \n$\\simplify{x^2-{a[0]}x+{a[0]}x-{a[0]*a[0]}}$ | \n(use the distributive law on each bracket) | \n
| \n | $=$ | \n$\\simplify{x^2-{a[0]*a[0]}}$ | \n(collect like terms) | \n
Method 2 (FOIL)
\nMultiply the First terms in each bracket, then the Outer terms, then the Inner terms and then the Last terms. Add them all together.
\n| $\\simplify{(x+{a[0]})(x-{a[0]})}$ | \n$=$ | \n\n $\\simplify[basic]{x^2-{a[0]}x+{a[0]}x-{a[0]*a[0]}}$ \n | \n\n (First, Outer, Inner, Last) \n | \n
| \n | $=$ | \n$\\simplify{x^2-{a[0]*a[0]}}$ | \n(collect like terms) | \n
Method 3 (difference of two squares)
\nNotice that the product will expand to be a difference of two squares. Square the first term minus the square of the second term.
\n| $\\simplify{(x+{a[0]})(x-{a[0]})}$ | \n$=$ | \n\n $\\simplify{x^2-{a[0]*a[0]}}$ \n | \n\n (difference of two squares) \n | \n
$\\simplify{(x+{a[2]})^2}$ = [[0]]
\n\n", "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "gaps": [{"notallowed": {"message": "Ensure you don't use brackets in your answer.
", "showStrings": false, "strings": ["(", ")"], "partialCredit": 0}, "vsetrangepoints": 5, "expectedvariablenames": ["x"], "checkingaccuracy": 0.001, "vsetrange": [0, 1], "showpreview": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "scripts": {}, "answer": "x^2+{2*a[2]}x+{a[2]*a[2]}", "marks": 1, "checkvariablenames": true, "checkingtype": "absdiff", "type": "jme"}], "steps": [{"prompt": "It is important to realise that $\\simplify{(x+{a[2]})^2}=\\simplify{(x+{a[2]})(x+{a[2]})}$. Recall that squaring something is multiplying it by itself.
\n\n
Method 1 (the distributive law)
\nWe expand $\\simplify{(x+{a[2]})(x+{a[2]})}$ one bracket at a time.
\n| $\\simplify{(x+{a[2]})(x+{a[2]})}$ | \n$=$ | \n\n $\\simplify{x(x+{a[2]})+{a[2]}(x+{a[2]})}$ \n | \n\n (each term in one bracket times the entire other bracket) \n | \n
| \n | $=$ | \n$\\simplify{x^2+{a[2]}x+{a[2]}x+{a[2]*a[2]}}$ | \n(use the distributive law on each bracket) | \n
| \n | $=$ | \n$\\simplify{x^2+{2*a[2]}x+{a[2]*a[2]}}$ | \n(collect like terms) | \n
Method 2 (FOIL)
\nMultiply the First terms in each bracket, then the Outer terms, then the Inner terms and then the Last terms. Add them all together.
\n| $\\simplify{(x+{a[2]})(x+{a[2]})}$ | \n$=$ | \n\n $\\simplify[basic]{x^2+{a[2]}x+{a[2]}x+{a[2]*a[2]}}$ \n | \n\n (First, Outer, Inner, Last) \n | \n
| \n | $=$ | \n$\\simplify{x^2+{2*a[2]}x+{a[2]*a[2]}}$ | \n(collect like terms) | \n
Method 3 (perfect square)
\nNotice that $\\simplify{(x+{a[2]})^2}$ is a perfect square. Square the first term, double the second term times the first, then square the last term, add them all together.
\n| $\\simplify{(x+{a[2]})}$ | \n$=$ | \n\n $\\simplify{x^2+{2*a[2]}x+{a[2]*a[2]}}$ \n | \n\n (perfect square) \n | \n
$\\simplify{(w+{a[1]})(w-{a[1]})}$ = [[0]]
\n\n", "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "gaps": [{"notallowed": {"message": "Ensure you don't use brackets in your answer.
", "showStrings": false, "strings": ["(", ")"], "partialCredit": 0}, "vsetrangepoints": 5, "expectedvariablenames": ["w"], "checkingaccuracy": 0.001, "vsetrange": [0, 1], "showpreview": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "scripts": {}, "answer": "w^2-{a[1]*a[1]}", "marks": 1, "checkvariablenames": true, "checkingtype": "absdiff", "type": "jme"}], "steps": [{"prompt": "Method 1 (the distributive law)
\nWe expand $\\simplify{(w+{a[1]})(w-{a[1]})}$ one bracket at a time.
\n| $\\simplify{(w+{a[1]})(w-{a[1]})}$ | \n$=$ | \n\n $\\simplify{w(w-{a[1]})+{a[1]}(w-{a[1]})}$ \n | \n\n (each term in one bracket times the entire other bracket) \n | \n
| \n | $=$ | \n$\\simplify{w^2-{a[1]}w+{a[1]}w-{a[1]*a[1]}}$ | \n(use the distributive law on each bracket) | \n
| \n | $=$ | \n$\\simplify{w^2-{a[1]*a[1]}}$ | \n(collect like terms) | \n
Method 2 (FOIL)
\nMultiply the First terms in each bracket, then the Outer terms, then the Inner terms and then the Last terms. Add them all together.
\n| $\\simplify{(w+{a[1]})(w-{a[1]})}$ | \n$=$ | \n\n $\\simplify[basic]{w^2-{a[1]}w+{a[1]}w-{a[1]*a[1]}}$ \n | \n\n (First, Outer, Inner, Last) \n | \n
| \n | $=$ | \n$\\simplify{w^2-{a[1]*a[1]}}$ | \n(collect like terms) | \n
Method 3 (difference of two squares)
\nNotice that the product will expand to be a difference of two squares. Square the first term minus the square of the second term.
\n| $\\simplify{(w+{a[1]})(w-{a[1]})}$ | \n$=$ | \n\n $\\simplify{w^2-{a[1]*a[1]}}$ \n | \n\n (difference of two squares) \n | \n
$\\simplify{(r+{a[3]})(r+{a[3]})}$ = [[0]]
\n\n", "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "gaps": [{"notallowed": {"message": "Ensure you don't use brackets in your answer.
", "showStrings": false, "strings": ["(", ")"], "partialCredit": 0}, "vsetrangepoints": 5, "expectedvariablenames": ["r"], "checkingaccuracy": 0.001, "vsetrange": [0, 1], "showpreview": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "scripts": {}, "answer": "r^2+{2*a[3]}r+{a[3]*a[3]}", "marks": 1, "checkvariablenames": true, "checkingtype": "absdiff", "type": "jme"}], "steps": [{"prompt": "Method 1 (the distributive law)
\nWe expand $\\simplify{(r+{a[3]})(r+{a[3]})}$ one bracket at a time.
\n| $\\simplify{(r+{a[3]})(r+{a[3]})}$ | \n$=$ | \n\n $\\simplify{r(r+{a[3]})+{a[3]}(r+{a[3]})}$ \n | \n\n (each term in one bracket times the entire other bracket) \n | \n
| \n | $=$ | \n$\\simplify{r^2+{a[3]}r+{a[3]}r+{a[3]*a[3]}}$ | \n(use the distributive law on each bracket) | \n
| \n | $=$ | \n$\\simplify{r^2+{2*a[3]}r+{a[3]*a[3]}}$ | \n(collect like terms) | \n
Method 2 (FOIL)
\nMultiply the First terms in each bracket, then the Outer terms, then the Inner terms and then the Last terms. Add them all together.
\n| $\\simplify{(r+{a[3]})(r+{a[3]})}$ | \n$=$ | \n\n $\\simplify[basic]{r^2+{a[3]}r+{a[3]}r+{a[3]*a[3]}}$ \n | \n\n (First, Outer, Inner, Last) \n | \n
| \n | $=$ | \n$\\simplify{r^2+{2*a[3]}r+{a[3]*a[3]}}$ | \n(collect like terms) | \n
Method 3 (perfect square)
\nNotice that $\\simplify{(r+{a[3]})(r+{a[3]})}$ is a perfect square. Square the first term, double the second term times the first, then square the last term, add them all together.
\n| $\\simplify{(r+{a[3]})(r+{a[3]})}$ | \n$=$ | \n\n $\\simplify{r^2+{2*a[3]}r+{a[3]*a[3]}}$ \n | \n\n (perfect square) \n | \n
Expand and simplify the following.
", "variable_groups": [], "variablesTest": {"maxRuns": "127", "condition": ""}, "variables": {"a": {"definition": "shuffle(-12..12 except 0)[0..4]", "templateType": "anything", "group": "Ungrouped variables", "name": "a", "description": ""}}, "metadata": {"notes": "", "description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "type": "question", "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "Expanding a binomial product (monic factors)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ben Brawn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/605/"}], "ungrouped_variables": ["a", "b"], "tags": ["binomial", "binomial product", "distributive law", "expanding", "Factorisation", "factorisation", "Factors", "factors", "monic", "quadratic"], "statement": "Expand and simplify the following.
", "preamble": {"css": "", "js": ""}, "functions": {}, "advice": "", "metadata": {"licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International", "description": ""}, "variables": {"b": {"templateType": "anything", "name": "b", "group": "Ungrouped variables", "definition": "shuffle(-12..12 except 0)[0..4]", "description": ""}, "a": {"templateType": "anything", "name": "a", "group": "Ungrouped variables", "definition": "shuffle(-12..12 except 0)[0..4]", "description": ""}}, "variablesTest": {"condition": "", "maxRuns": "127"}, "rulesets": {}, "parts": [{"marks": 0, "prompt": "$\\simplify{(x+{a[0]})(x+{b[0]})}$ = [[0]]
\n\n", "gaps": [{"notallowed": {"strings": ["(", ")"], "partialCredit": 0, "message": "Ensure you don't use brackets in your answer.
", "showStrings": false}, "marks": 1, "unitTests": [], "showCorrectAnswer": true, "failureRate": 1, "variableReplacements": [], "showFeedbackIcon": true, "vsetRange": [0, 1], "checkingType": "absdiff", "scripts": {}, "checkingAccuracy": 0.001, "variableReplacementStrategy": "originalfirst", "showPreview": true, "type": "jme", "answer": "x^2+{a[0]+b[0]}x+{a[0]*b[0]}", "customMarkingAlgorithm": "", "vsetRangePoints": 5, "checkVariableNames": true, "expectedVariableNames": ["x"], "extendBaseMarkingAlgorithm": true}], "unitTests": [], "steps": [{"marks": 0, "prompt": "Method 1 (the distributive law)
\nWe expand $\\simplify{(x+{a[0]})(x+{b[0]})}$ one bracket at a time.
\n| $\\simplify{(x+{a[0]})(x+{b[0]})}$ | \n$=$ | \n\n $\\simplify{x(x+{b[0]})+{a[0]}(x+{b[0]})}$ \n | \n\n (each term in one bracket times the entire other bracket) \n | \n
| \n | $=$ | \n$\\simplify{x^2+{b[0]}x+{a[0]}x+{a[0]*b[0]}}$ | \n(use the distributive law on each bracket) | \n
| \n | $=$ | \n$\\simplify{x^2+{b[0]+a[0]}x+{a[0]*b[0]}}$ | \n(collect like terms) | \n
Method 2 (FOIL)
\nMultiply the First terms in each bracket, then the Outer terms, then the Inner terms and then the Last terms. Add them all together.
\n| $\\simplify{(x+{a[0]})(x+{b[0]})}$ | \n$=$ | \n\n $\\simplify[basic]{x^2+{b[0]}x+{a[0]}x+{a[0]*b[0]}}$ \n | \n\n (First, Outer, Inner, Last) \n | \n
| \n | $=$ | \n$\\simplify{x^2+{b[0]+a[0]}x+{a[0]*b[0]}}$ | \n(collect like terms) | \n
$\\simplify{(x+{a[1]})(x+{b[1]})}$ = [[0]]
\n\n", "gaps": [{"notallowed": {"strings": ["(", ")"], "partialCredit": 0, "message": "Ensure you don't use brackets in your answer.
", "showStrings": false}, "marks": 1, "unitTests": [], "showCorrectAnswer": true, "failureRate": 1, "variableReplacements": [], "showFeedbackIcon": true, "vsetRange": [0, 1], "checkingType": "absdiff", "scripts": {}, "checkingAccuracy": 0.001, "variableReplacementStrategy": "originalfirst", "showPreview": true, "type": "jme", "answer": "x^2+{a[1]+b[1]}x+{a[1]*b[1]}", "customMarkingAlgorithm": "", "vsetRangePoints": 5, "checkVariableNames": true, "expectedVariableNames": ["x"], "extendBaseMarkingAlgorithm": true}], "unitTests": [], "steps": [{"marks": 0, "prompt": "Method 1 (the distributive law)
\nWe expand $\\simplify{(x+{a[1]})(x+{b[1]})}$ one bracket at a time.
\n| $\\simplify{(x+{a[1]})(x+{b[1]})}$ | \n$=$ | \n\n $\\simplify{x(x+{b[1]})+{a[1]}(x+{b[1]})}$ \n | \n\n (each term in one bracket times the entire other bracket) \n | \n
| \n | $=$ | \n$\\simplify{x^2+{b[1]}x+{a[1]}x+{a[1]*b[1]}}$ | \n(use the distributive law on each bracket) | \n
| \n | $=$ | \n$\\simplify{x^2+{b[1]+a[1]}x+{a[1]*b[1]}}$ | \n(collect like terms) | \n
Method 2 (FOIL)
\nMultiply the First terms in each bracket, then the Outer terms, then the Inner terms and then the Last terms. Add them all together.
\n| $\\simplify{(x+{a[1]})(x+{b[1]})}$ | \n$=$ | \n\n $\\simplify[basic]{x^2+{b[1]}x+{a[1]}x+{a[1]*b[1]}}$ \n | \n\n (First, Outer, Inner, Last) \n | \n
| \n | $=$ | \n$\\simplify{x^2+{b[1]+a[1]}x+{a[1]*b[1]}}$ | \n(collect like terms) | \n
$\\simplify{(m+{a[2]})(m+{b[2]})}$ = [[0]]
\n\n", "gaps": [{"notallowed": {"strings": ["(", ")"], "partialCredit": 0, "message": "Ensure you don't use brackets in your answer.
", "showStrings": false}, "marks": 1, "unitTests": [], "showCorrectAnswer": true, "failureRate": 1, "variableReplacements": [], "showFeedbackIcon": true, "vsetRange": [0, 1], "checkingType": "absdiff", "scripts": {}, "checkingAccuracy": 0.001, "variableReplacementStrategy": "originalfirst", "showPreview": true, "type": "jme", "answer": "m^2+{a[2]+b[2]}m+{a[2]*b[2]}", "customMarkingAlgorithm": "", "vsetRangePoints": 5, "checkVariableNames": true, "expectedVariableNames": ["m"], "extendBaseMarkingAlgorithm": true}], "unitTests": [], "steps": [{"marks": 0, "prompt": "Method 1 (the distributive law)
\nWe expand $\\simplify{(m+{a[2]})(m+{b[2]})}$ one bracket at a time.
\n| $\\simplify{(m+{a[2]})(m+{b[2]})}$ | \n$=$ | \n\n $\\simplify{m(m+{b[2]})+{a[2]}(m+{b[2]})}$ \n | \n\n (each term in one bracket times the entire other bracket) \n | \n
| \n | $=$ | \n$\\simplify{m^2+{b[2]}m+{a[2]}m+{a[2]*b[2]}}$ | \n(use the distributive law on each bracket) | \n
| \n | $=$ | \n$\\simplify{m^2+{b[2]+a[2]}m+{a[2]*b[2]}}$ | \n(collect like terms) | \n
Method 2 (FOIL)
\nMultiply the First terms in each bracket, then the Outer terms, then the Inner terms and then the Last terms. Add them all together.
\n| $\\simplify{(m+{a[2]})(m+{b[2]})}$ | \n$=$ | \n\n $\\simplify[basic]{m^2+{b[2]}m+{a[2]}m+{a[2]*b[2]}}$ \n | \n\n (First, Outer, Inner, Last) \n | \n
| \n | $=$ | \n$\\simplify{m^2+{b[2]+a[2]}m+{a[2]*b[2]}}$ | \n(collect like terms) | \n
$\\simplify{(t+{a[3]})(t+{b[3]})}$ = [[0]]
\n\n", "gaps": [{"notallowed": {"strings": ["(", ")"], "partialCredit": 0, "message": "Ensure you don't use brackets in your answer.
", "showStrings": false}, "marks": 1, "unitTests": [], "showCorrectAnswer": true, "failureRate": 1, "variableReplacements": [], "showFeedbackIcon": true, "vsetRange": [0, 1], "checkingType": "absdiff", "scripts": {}, "checkingAccuracy": 0.001, "variableReplacementStrategy": "originalfirst", "showPreview": true, "type": "jme", "answer": "t^2+{a[3]+b[3]}t+{a[3]*b[3]}", "customMarkingAlgorithm": "", "vsetRangePoints": 5, "checkVariableNames": true, "expectedVariableNames": ["t"], "extendBaseMarkingAlgorithm": true}], "unitTests": [], "steps": [{"marks": 0, "prompt": "Method 1 (the distributive law)
\nWe expand $\\simplify{(t+{a[3]})(t+{b[3]})}$ one bracket at a time.
\n| $\\simplify{(t+{a[3]})(t+{b[3]})}$ | \n$=$ | \n\n $\\simplify{t(t+{b[3]})+{a[3]}(t+{b[3]})}$ \n | \n\n (each term in one bracket times the entire other bracket) \n | \n
| \n | $=$ | \n$\\simplify{t^2+{b[3]}t+{a[3]}t+{a[3]*b[3]}}$ | \n(use the distributive law on each bracket) | \n
| \n | $=$ | \n$\\simplify{t^2+{b[3]+a[3]}t+{a[3]*b[3]}}$ | \n\n\n (collect like terms) \n | \n
Method 2 (FOIL)
\nMultiply the First terms in each bracket, then the Outer terms, then the Inner terms and then the Last terms. Add them all together.
\n| $\\simplify{(t+{a[3]})(t+{b[3]})}$ | \n$=$ | \n\n $\\simplify[basic]{t^2+{b[3]}t+{a[3]}t+{a[3]*b[3]}}$ \n | \n\n (First, Outer, Inner, Last) \n | \n
| \n | $=$ | \n$\\simplify{t^2+{b[3]+a[3]}t+{a[3]*b[3]}}$ | \n(collect like terms) | \n
$\\simplify[basic]{({c[0]}x+{a[0]})({d[0]}x+{b[0]})}$ = [[0]]
\n\n", "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "gaps": [{"notallowed": {"message": "Ensure you don't use brackets in your answer.
", "showStrings": false, "strings": ["(", ")"], "partialCredit": 0}, "vsetrangepoints": 5, "expectedvariablenames": ["x"], "checkingaccuracy": 0.001, "vsetrange": [0, 1], "showpreview": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "scripts": {}, "answer": "{c[0]*d[0]}x^2+{d[0]*a[0]+c[0]*b[0]}x+{a[0]*b[0]}", "marks": 1, "checkvariablenames": true, "checkingtype": "absdiff", "type": "jme"}], "steps": [{"prompt": "Method 1 (the distributive law)
\nWe expand $\\simplify[basic]{({c[0]}x+{a[0]})({d[0]}x+{b[0]})}$ one bracket at a time.
\n| $\\simplify[basic]{({c[0]}x+{a[0]})({d[0]}x+{b[0]})}$ | \n$=$ | \n\n $\\simplify[basic]{{c[0]}x({d[0]}x+{b[0]})+{a[0]}({d[0]}x+{b[0]})}$ \n | \n\n (each term in one bracket times the entire other bracket) \n | \n
| \n | $=$ | \n$\\simplify[basic]{{c[0]*d[0]}x^2+{c[0]*b[0]}x+{d[0]*a[0]}x+{a[0]*b[0]}}$ | \n(use the distributive law on each bracket) | \n
| \n | $=$ | \n$\\simplify[basic]{{c[0]*d[0]}x^2+{d[0]*a[0]+c[0]*b[0]}x+{a[0]*b[0]}}$ | \n(collect like terms) | \n
Method 2 (FOIL)
\nMultiply the First terms in each bracket, then the Outer terms, then the Inner terms and then the Last terms. Add them all together.
\n| $\\simplify[basic]{({c[0]}x+{a[0]})({d[0]}x+{b[0]})}$ | \n$=$ | \n\n $\\simplify[basic]{{c[0]*d[0]}x^2+{c[0]*b[0]}x+{d[0]*a[0]}x+{a[0]*b[0]}}$ \n | \n\n (First, Outer, Inner, Last) \n | \n
| \n | $=$ | \n$\\simplify[basic]{{c[0]*d[0]}x^2+{d[0]*a[0]+c[0]*b[0]}x+{a[0]*b[0]}}$ | \n(collect like terms) | \n
$\\simplify[basic]{({c[1]}x+{a[1]})({d[1]}x+{b[1]})}$ = [[0]]
\n\n", "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "gaps": [{"notallowed": {"message": "Ensure you don't use brackets in your answer.
", "showStrings": false, "strings": ["(", ")"], "partialCredit": 0}, "vsetrangepoints": 5, "expectedvariablenames": ["x"], "checkingaccuracy": 0.001, "vsetrange": [0, 1], "showpreview": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "scripts": {}, "answer": "{c[1]*d[1]}x^2+{d[1]*a[1]+c[1]*b[1]}x+{a[1]*b[1]}", "marks": 1, "checkvariablenames": true, "checkingtype": "absdiff", "type": "jme"}], "steps": [{"prompt": "Method 1 (the distributive law)
\nWe expand $\\simplify[basic]{({c[1]}x+{a[1]})({d[1]}x+{b[1]})}$ one bracket at a time.
\n| $\\simplify[basic]{({c[1]}x+{a[1]})({d[1]}x+{b[1]})}$ | \n$=$ | \n\n $\\simplify[basic]{{c[1]}x({d[1]}x+{b[1]})+{a[1]}({d[1]}x+{b[1]})}$ \n | \n\n (each term in one bracket times the entire other bracket) \n | \n
| \n | $=$ | \n$\\simplify[basic]{{c[1]*d[1]}x^2+{c[1]*b[1]}x+{d[1]*a[1]}x+{a[1]*b[1]}}$ | \n(use the distributive law on each bracket) | \n
| \n | $=$ | \n$\\simplify[basic]{{c[1]*d[1]}x^2+{d[1]*a[1]+c[1]*b[1]}x+{a[1]*b[1]}}$ | \n(collect like terms) | \n
Method 2 (FOIL)
\nMultiply the First terms in each bracket, then the Outer terms, then the Inner terms and then the Last terms. Add them all together.
\n| $\\simplify[basic]{({c[1]}x+{a[1]})({d[1]}x+{b[1]})}$ | \n$=$ | \n\n $\\simplify[basic]{{c[1]*d[1]}x^2+{c[1]*b[1]}x+{d[1]*a[1]}x+{a[1]*b[1]}}$ \n | \n\n (First, Outer, Inner, Last) \n | \n
| \n | $=$ | \n$\\simplify[basic]{{c[1]*d[1]}x^2+{d[1]*a[1]+c[1]*b[1]}x+{a[1]*b[1]}}$ | \n(collect like terms) | \n
Expand and simplify the following.
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\nThe first has two negative roots, the second has one negative and one positive, and the third is the difference of two squares.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Factorise the following quadratic equations.
\n", "variables": {"v1": {"name": "v1", "group": "Part A ", "definition": "random(1..10)", "templateType": "anything", "description": ""}, "v2": {"name": "v2", "group": "Part A ", "definition": "random(2..6 except v1)", "templateType": "anything", "description": ""}, "v4": {"name": "v4", "group": "Part A ", "definition": "random(1..10 except -v3)", "templateType": "anything", "description": ""}, "v5": {"name": "v5", "group": "Part A ", "definition": "random(2..10)", "templateType": "anything", "description": ""}, "v3": {"name": "v3", "group": "Part A ", "definition": "random(-8..-1)", "templateType": "anything", "description": ""}, "v6": {"name": "v6", "group": "Part A ", "definition": "-v5", "templateType": "anything", "description": ""}}, "tags": ["Factorisation", "factorisation", "factorising quadratic equations", "Factorising quadratic equations", "taxonomy"], "ungrouped_variables": [], "functions": {}, "preamble": {"js": "question.is_factorised = function(part,penalty) {\n penalty = penalty || 0;\n if(part.credit>0) {\n // Parse the student's answer as a syntax tree\n var studentTree = Numbas.jme.compile(part.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_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 part.multCredit(penalty,'Your answer is not fully factorised.');\n }\n }\n}", "css": ""}, "advice": "Quadratic equations of the form
\n\\[x^2+bx+c=0\\]
\ncan be factorised to create an equation of the form
\n\\[(x+m)(x+n)=0\\text{.}\\]
\nWhen we expand a factorised quadratic expression we obtain
\n\\[(x+m)(x+n)=x^2+(m+n)x+(m \\times n)\\text{.}\\]
\nTo factorise an equation of the form $x^2+bx+c$, we need to find two numbers which add together to make $b$, and multiply together to make $c$.
\n\\[\\simplify{x^2+{v1+v2}x+{v1*v2}=0}\\]
\nWe need to find two values that add together to make $\\var{v1+v2}$ and multiply together to make $\\var{v1*v2}$.
\n\\[\\begin{align}
\\var{v1} \\times \\var{v2}&=\\var{v1*v2}\\\\
\\var{v1}+\\var{v2}&=\\var{v1+v2}\\\\
\\end{align} \\]
So the factorised form of the equation is
\n\\[\\simplify{(x+{v1})(x+{v2})}=0\\text{.}\\]
\n\nWe can begin factorising by finding factors of $\\var{v3*v4}$ that add together to give $\\var{v3+v4}$.
\n\\[\\begin{align}
\\var{v3} \\times \\var{v4}&=\\var{v3*v4}\\\\
\\var{v3}+\\var{v4}&=\\var{v3+v4}\\\\
\\end{align} \\]
So the factorised form of the equation is
\n\\[\\simplify{(x+{v3})(x+{v4})}=0\\text{.}\\]
\nWhen factorising the quadratic expression
\n\\[\\simplify{x^2+{v5*v6}=0}\\]
\nwe need to find two values that add together to make $0$ and multiply together to make $\\var{v5*v6}$.
\n\\begin{align}
\\var{v5} \\times \\var{v6}& = \\var{v5*v6}\\\\
\\simplify[]{ {v5} + {v6}} &= 0 \\\\
\\end{align}
So the factorised form of the equation is
\n\\[\\simplify{(x+{v5})(x+{v6})}=0\\text{.}\\]
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\n[[0]] $=0$
\n", "variableReplacementStrategy": "originalfirst"}, {"scripts": {}, "variableReplacements": [], "customName": "", "useCustomName": false, "unitTests": [], "extendBaseMarkingAlgorithm": true, "showFeedbackIcon": true, "sortAnswers": false, "showCorrectAnswer": true, "gaps": [{"mustmatchpattern": {"pattern": "(`+-x^$n`? + `+- $n)`* * $z", "message": "Your answer is not fully factorised.", "partialCredit": 0, "nameToCompare": ""}, "variableReplacementStrategy": "originalfirst", "unitTests": [], "checkingAccuracy": 0.001, "scripts": {}, "failureRate": 1, "checkVariableNames": false, "marks": 1, "valuegenerators": [{"name": "x", "value": ""}], "variableReplacements": [], "useCustomName": false, "showPreview": true, "customName": "", "extendBaseMarkingAlgorithm": true, "checkingType": "absdiff", "vsetRange": [0, 1], "showCorrectAnswer": true, "type": "jme", "answer": "(x+{v3})(x+{v4})", "customMarkingAlgorithm": "", "showFeedbackIcon": true, "vsetRangePoints": 5}], "type": "gapfill", "marks": 0, "customMarkingAlgorithm": "", "prompt": "
$\\simplify{x^2+{v3+v4}x+{v3*v4}}=0$
\n[[0]] $=0$
\n", "variableReplacementStrategy": "originalfirst"}, {"scripts": {}, "variableReplacements": [], "customName": "", "useCustomName": false, "unitTests": [], "extendBaseMarkingAlgorithm": true, "showFeedbackIcon": true, "sortAnswers": false, "showCorrectAnswer": true, "gaps": [{"mustmatchpattern": {"pattern": "(`+-x^$n`? + `+- $n)`* * $z", "message": "
$\\simplify{x^2+{v5*v6}}=0$
\n[[0]] $=0$
", "variableReplacementStrategy": "originalfirst"}], "variablesTest": {"condition": "", "maxRuns": 100}, "rulesets": {}}, {"name": "Factorising Quadratic Equations with $x^2$ Coefficients Greater than 1", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Hannah Aldous", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1594/"}], "advice": "As this question involves a number greater than $1$ before the $x^2$ value it has a factorised form $(ax+b)(cx+d)$.
\nTo find $a$ and $c$, we need to consider the factors of $\\var{a*c}$.
\nWe are already given that one of them is $\\var{a}$, so we know that the other one must be $\\var{c}$.
\nThis means our factorised equation must take the form
\n\\[(\\var{a}x+b)(\\var{c}x+d)=0\\text{.}\\]
\nThis expands to
\n\\[ \\simplify{ {a*c}x^2 + ({a}*d+{c}*b)x + a*b} \\]
\nSo we must find two numbers which add together to make $\\var{a*d+b*c}$, and multiply together to make $\\var{b*d}$.
\nTherefore $b$ and $d$ must satisfy
\n\\begin{align}
b \\times d &=\\var{b*d}\\\\
\\simplify{{a}d+{c}b} &= \\var{a*d+b*c}\\text{.}
\\end{align}
$b = \\var{b}$ and $d = \\var{d}$ satisfy these equations:
\n\\begin{align}
\\var{b} \\times \\var{d} &=\\var{b*d}\\\\
\\simplify[]{ {a}*{d} + {b}*{c} } &= \\var{a*d+b*c}
\\end{align}
So the factorised form of the equation is
\n\\[ \\simplify{({a}x+{b})({c}x+{d}) = 0} \\text{.}\\]
\n$\\simplify{({a}x+{b})({c}x+{d}) = 0}$ when either $\\var{a}x+\\var{b} = 0$ or $\\var{c}x+ \\var{d} = 0$.
\nSo the roots of the equation are $\\var[fractionnumbers]{-b/a}$ and $\\var[fractionnumbers]{-d/c}$.
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\n$\\simplify{{a*c}x^2+{a*d+b*c}x+{b*d}=0}\\text{.}$
\n$(\\var{a}x+\\phantom{.}$[[0]]$) ($[[1]]$x+\\phantom{.}$[[2]]$)\\; = 0$
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\nInput your answer as $x_1$ and $x_2$, where $x_1<x_2$.
\n$x_1=$ [[0]]
\n$x_2=$ [[1]]
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