// Numbas version: exam_results_page_options {"name": "Expanding binomial products", "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "duration": 0, "percentPass": 0, "showQuestionGroupNames": false, "shuffleQuestionGroups": false, "showstudentname": true, "question_groups": [{"name": "Group", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": ["", "", ""], "variable_overrides": [[], [], []], "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.

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$\\simplify{(x+{a[0]})(x+{b[0]})}$ = [[0]]

", "gaps": [{"notallowed": {"strings": ["(", ")"], "partialCredit": 0, "message": "

Ensure you don't use brackets in your answer.

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Method 1 (the distributive law)

We expand $\\simplify{(x+{a[0]})(x+{b[0]})}$ one bracket at a time.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\simplify{(x+{a[0]})(x+{b[0]})}$	$=$	\n $\\simplify{x(x+{b[0]})+{a[0]}(x+{b[0]})}$ \n	\n (each term in one bracket times the entire other bracket) \n
	$=$	$\\simplify{x^2+{b[0]}x+{a[0]}x+{a[0]*b[0]}}$	(use the distributive law on each bracket)
	$=$	$\\simplify{x^2+{b[0]+a[0]}x+{a[0]*b[0]}}$	(collect like terms)

Method 2 (FOIL)

Multiply the First terms in each bracket, then the Outer terms, then the Inner terms and then the Last terms. Add them all together.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\simplify{(x+{a[0]})(x+{b[0]})}$	$=$	\n $\\simplify[basic]{x^2+{b[0]}x+{a[0]}x+{a[0]*b[0]}}$ \n	\n (First, Outer, Inner, Last) \n
	$=$	$\\simplify{x^2+{b[0]+a[0]}x+{a[0]*b[0]}}$	(collect like terms)

$\\simplify{(x+{a[1]})(x+{b[1]})}$ = [[0]]

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Ensure you don't use brackets in your answer.

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Method 1 (the distributive law)

We expand $\\simplify{(x+{a[1]})(x+{b[1]})}$ one bracket at a time.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\simplify{(x+{a[1]})(x+{b[1]})}$	$=$	\n $\\simplify{x(x+{b[1]})+{a[1]}(x+{b[1]})}$ \n	\n (each term in one bracket times the entire other bracket) \n
	$=$	$\\simplify{x^2+{b[1]}x+{a[1]}x+{a[1]*b[1]}}$	(use the distributive law on each bracket)
	$=$	$\\simplify{x^2+{b[1]+a[1]}x+{a[1]*b[1]}}$	(collect like terms)

Method 2 (FOIL)

Multiply the First terms in each bracket, then the Outer terms, then the Inner terms and then the Last terms. Add them all together.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\simplify{(x+{a[1]})(x+{b[1]})}$	$=$	\n $\\simplify[basic]{x^2+{b[1]}x+{a[1]}x+{a[1]*b[1]}}$ \n	\n (First, Outer, Inner, Last) \n
	$=$	$\\simplify{x^2+{b[1]+a[1]}x+{a[1]*b[1]}}$	(collect like terms)

$\\simplify{(m+{a[2]})(m+{b[2]})}$ = [[0]]

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Ensure you don't use brackets in your answer.

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Method 1 (the distributive law)

We expand $\\simplify{(m+{a[2]})(m+{b[2]})}$ one bracket at a time.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\simplify{(m+{a[2]})(m+{b[2]})}$	$=$	\n $\\simplify{m(m+{b[2]})+{a[2]}(m+{b[2]})}$ \n	\n (each term in one bracket times the entire other bracket) \n
	$=$	$\\simplify{m^2+{b[2]}m+{a[2]}m+{a[2]*b[2]}}$	(use the distributive law on each bracket)
	$=$	$\\simplify{m^2+{b[2]+a[2]}m+{a[2]*b[2]}}$	(collect like terms)

Method 2 (FOIL)

Multiply the First terms in each bracket, then the Outer terms, then the Inner terms and then the Last terms. Add them all together.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\simplify{(m+{a[2]})(m+{b[2]})}$	$=$	\n $\\simplify[basic]{m^2+{b[2]}m+{a[2]}m+{a[2]*b[2]}}$ \n	\n (First, Outer, Inner, Last) \n
	$=$	$\\simplify{m^2+{b[2]+a[2]}m+{a[2]*b[2]}}$	(collect like terms)

$\\simplify{(t+{a[3]})(t+{b[3]})}$ = [[0]]

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Ensure you don't use brackets in your answer.

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Method 1 (the distributive law)

We expand $\\simplify{(t+{a[3]})(t+{b[3]})}$ one bracket at a time.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\simplify{(t+{a[3]})(t+{b[3]})}$

$=$

$\\simplify{t(t+{b[3]})+{a[3]}(t+{b[3]})}$

(each term in one bracket times the entire other bracket)

$=$

$\\simplify{t^2+{b[3]}t+{a[3]}t+{a[3]*b[3]}}$

(use the distributive law on each bracket)

$=$

$\\simplify{t^2+{b[3]+a[3]}t+{a[3]*b[3]}}$

(collect like terms)

Method 2 (FOIL)

Multiply the First terms in each bracket, then the Outer terms, then the Inner terms and then the Last terms. Add them all together.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\simplify{(t+{a[3]})(t+{b[3]})}$	$=$	\n $\\simplify[basic]{t^2+{b[3]}t+{a[3]}t+{a[3]*b[3]}}$ \n	\n (First, Outer, Inner, Last) \n
	$=$	$\\simplify{t^2+{b[3]+a[3]}t+{a[3]*b[3]}}$	(collect like terms)

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$\\simplify{(x+{a[0]})(x-{a[0]})}$ = [[0]]

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Ensure you don't use brackets in your answer.

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Method 1 (the distributive law)

We expand $\\simplify{(x+{a[0]})(x-{a[0]})}$ one bracket at a time.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\simplify{(x+{a[0]})(x-{a[0]})}$	$=$	\n $\\simplify{x(x-{a[0]})+{a[0]}(x-{a[0]})}$ \n	\n (each term in one bracket times the entire other bracket) \n
	$=$	$\\simplify{x^2-{a[0]}x+{a[0]}x-{a[0]*a[0]}}$	(use the distributive law on each bracket)
	$=$	$\\simplify{x^2-{a[0]*a[0]}}$	(collect like terms)

Method 2 (FOIL)

Multiply the First terms in each bracket, then the Outer terms, then the Inner terms and then the Last terms. Add them all together.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\simplify{(x+{a[0]})(x-{a[0]})}$	$=$	\n $\\simplify[basic]{x^2-{a[0]}x+{a[0]}x-{a[0]*a[0]}}$ \n	\n (First, Outer, Inner, Last) \n
	$=$	$\\simplify{x^2-{a[0]*a[0]}}$	(collect like terms)

Method 3 (difference of two squares)

Notice that the product will expand to be a difference of two squares. Square the first term minus the square of the second term.

\n\n\n\n\n\n\n\n\n\n

$\\simplify{(x+{a[0]})(x-{a[0]})}$

$=$

$\\simplify{x^2-{a[0]*a[0]}}$

(difference of two squares)

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$\\simplify{(x+{a[2]})^2}$ = [[0]]

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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.

Method 1 (the distributive law)

We expand $\\simplify{(x+{a[2]})(x+{a[2]})}$ one bracket at a time.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\simplify{(x+{a[2]})(x+{a[2]})}$	$=$	\n $\\simplify{x(x+{a[2]})+{a[2]}(x+{a[2]})}$ \n	\n (each term in one bracket times the entire other bracket) \n
	$=$	$\\simplify{x^2+{a[2]}x+{a[2]}x+{a[2]*a[2]}}$	(use the distributive law on each bracket)
	$=$	$\\simplify{x^2+{2a[2]}x+{a[2]a[2]}}$	(collect like terms)

Method 2 (FOIL)

Multiply the First terms in each bracket, then the Outer terms, then the Inner terms and then the Last terms. Add them all together.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\simplify{(x+{a[2]})(x+{a[2]})}$	$=$	\n $\\simplify[basic]{x^2+{a[2]}x+{a[2]}x+{a[2]*a[2]}}$ \n	\n (First, Outer, Inner, Last) \n
	$=$	$\\simplify{x^2+{2a[2]}x+{a[2]a[2]}}$	(collect like terms)

Method 3 (perfect square)

Notice 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\n\n\n\n\n\n\n\n\n

$\\simplify{(x+{a[2]})}$

$=$

$\\simplify{x^2+{2*a[2]}x+{a[2]*a[2]}}$

(perfect square)

$\\simplify{(w+{a[1]})(w-{a[1]})}$ = [[0]]

", "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)

We expand $\\simplify{(w+{a[1]})(w-{a[1]})}$ one bracket at a time.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\simplify{(w+{a[1]})(w-{a[1]})}$	$=$	\n $\\simplify{w(w-{a[1]})+{a[1]}(w-{a[1]})}$ \n	\n (each term in one bracket times the entire other bracket) \n
	$=$	$\\simplify{w^2-{a[1]}w+{a[1]}w-{a[1]*a[1]}}$	(use the distributive law on each bracket)
	$=$	$\\simplify{w^2-{a[1]*a[1]}}$	(collect like terms)

Method 2 (FOIL)

Multiply the First terms in each bracket, then the Outer terms, then the Inner terms and then the Last terms. Add them all together.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\simplify{(w+{a[1]})(w-{a[1]})}$	$=$	\n $\\simplify[basic]{w^2-{a[1]}w+{a[1]}w-{a[1]*a[1]}}$ \n	\n (First, Outer, Inner, Last) \n
	$=$	$\\simplify{w^2-{a[1]*a[1]}}$	(collect like terms)

Method 3 (difference of two squares)

Notice that the product will expand to be a difference of two squares. Square the first term minus the square of the second term.

\n\n\n\n\n\n\n\n\n\n

$\\simplify{(w+{a[1]})(w-{a[1]})}$

$=$

$\\simplify{w^2-{a[1]*a[1]}}$

(difference of two squares)

$\\simplify{(r+{a[3]})(r+{a[3]})}$ = [[0]]

", "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)

We expand $\\simplify{(r+{a[3]})(r+{a[3]})}$ one bracket at a time.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\simplify{(r+{a[3]})(r+{a[3]})}$	$=$	\n $\\simplify{r(r+{a[3]})+{a[3]}(r+{a[3]})}$ \n	\n (each term in one bracket times the entire other bracket) \n
	$=$	$\\simplify{r^2+{a[3]}r+{a[3]}r+{a[3]*a[3]}}$	(use the distributive law on each bracket)
	$=$	$\\simplify{r^2+{2a[3]}r+{a[3]a[3]}}$	(collect like terms)

Method 2 (FOIL)

Multiply the First terms in each bracket, then the Outer terms, then the Inner terms and then the Last terms. Add them all together.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\simplify{(r+{a[3]})(r+{a[3]})}$	$=$	\n $\\simplify[basic]{r^2+{a[3]}r+{a[3]}r+{a[3]*a[3]}}$ \n	\n (First, Outer, Inner, Last) \n
	$=$	$\\simplify{r^2+{2a[3]}r+{a[3]a[3]}}$	(collect like terms)

Method 3 (perfect square)

Notice 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\n\n\n\n\n\n\n\n\n

$\\simplify{(r+{a[3]})(r+{a[3]})}$

$=$

$\\simplify{r^2+{2*a[3]}r+{a[3]*a[3]}}$

(perfect square)

Expand and simplify the following.

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$\\simplify[basic]{({c[0]}x+{a[0]})({d[0]}x+{b[0]})}$ = [[0]]

", "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "gaps": [{"notallowed": {"message": "

Ensure you don't use brackets in your answer.

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Method 1 (the distributive law)

We expand $\\simplify[basic]{({c[0]}x+{a[0]})({d[0]}x+{b[0]})}$ one bracket at a time.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\simplify[basic]{({c[0]}x+{a[0]})({d[0]}x+{b[0]})}$	$=$	\n $\\simplify[basic]{{c[0]}x({d[0]}x+{b[0]})+{a[0]}({d[0]}x+{b[0]})}$ \n	\n (each term in one bracket times the entire other bracket) \n
	$=$	$\\simplify[basic]{{c[0]d[0]}x^2+{c[0]b[0]}x+{d[0]a[0]}x+{a[0]b[0]}}$	(use the distributive law on each bracket)
	$=$	$\\simplify[basic]{{c[0]d[0]}x^2+{d[0]a[0]+c[0]b[0]}x+{a[0]b[0]}}$	(collect like terms)

Method 2 (FOIL)

Multiply the First terms in each bracket, then the Outer terms, then the Inner terms and then the Last terms. Add them all together.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\simplify[basic]{({c[0]}x+{a[0]})({d[0]}x+{b[0]})}$	$=$	\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 (First, Outer, Inner, Last) \n
	$=$	$\\simplify[basic]{{c[0]d[0]}x^2+{d[0]a[0]+c[0]b[0]}x+{a[0]b[0]}}$	(collect like terms)

$\\simplify[basic]{({c[1]}x+{a[1]})({d[1]}x+{b[1]})}$ = [[0]]

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Ensure you don't use brackets in your answer.

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Method 1 (the distributive law)

We expand $\\simplify[basic]{({c[1]}x+{a[1]})({d[1]}x+{b[1]})}$ one bracket at a time.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\simplify[basic]{({c[1]}x+{a[1]})({d[1]}x+{b[1]})}$	$=$	\n $\\simplify[basic]{{c[1]}x({d[1]}x+{b[1]})+{a[1]}({d[1]}x+{b[1]})}$ \n	\n (each term in one bracket times the entire other bracket) \n
	$=$	$\\simplify[basic]{{c[1]d[1]}x^2+{c[1]b[1]}x+{d[1]a[1]}x+{a[1]b[1]}}$	(use the distributive law on each bracket)
	$=$	$\\simplify[basic]{{c[1]d[1]}x^2+{d[1]a[1]+c[1]b[1]}x+{a[1]b[1]}}$	(collect like terms)

Method 2 (FOIL)

Multiply the First terms in each bracket, then the Outer terms, then the Inner terms and then the Last terms. Add them all together.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\simplify[basic]{({c[1]}x+{a[1]})({d[1]}x+{b[1]})}$	$=$	\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 (First, Outer, Inner, Last) \n
	$=$	$\\simplify[basic]{{c[1]d[1]}x^2+{d[1]a[1]+c[1]b[1]}x+{a[1]b[1]}}$	(collect like terms)

Expand and simplify the following.

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