$\\simplify{x^2+{v1+v2}x+{v1*v2}=0}$
\n[[0]] $=0$
\n", "variableReplacementStrategy": "originalfirst", "marks": 0, "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": []}, {"gaps": [{"scripts": {"mark": {"order": "after", "script": "question.is_factorised(this);"}}, "checkingaccuracy": 0.001, "marks": 1, "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "showpreview": true, "checkvariablenames": false, "checkingtype": "absdiff", "expectedvariablenames": [], "type": "jme", "answer": "(x+{v3})(x+{v4})", "vsetrange": [0, 1], "showCorrectAnswer": true, "vsetrangepoints": 5, "variableReplacements": []}], "scripts": {}, "type": "gapfill", "prompt": "
$\\simplify{x^2+{v3+v4}x+{v3*v4}}=0$
\n[[0]] $=0$
\n", "variableReplacementStrategy": "originalfirst", "marks": 0, "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": []}, {"gaps": [{"scripts": {"mark": {"order": "after", "script": "question.is_factorised(this);"}}, "checkingaccuracy": 0.001, "marks": 1, "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "showpreview": true, "checkvariablenames": false, "checkingtype": "absdiff", "expectedvariablenames": [], "type": "jme", "answer": "(x+{v5})(x+{v6})", "vsetrange": [0, 1], "showCorrectAnswer": true, "vsetrangepoints": 5, "variableReplacements": []}], "scripts": {}, "type": "gapfill", "prompt": "
$\\simplify{x^2+{v5*v6}}=0$
\n[[0]] $=0$
", "variableReplacementStrategy": "originalfirst", "marks": 0, "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": []}], "variable_groups": [{"variables": ["v1", "v2", "v3", "v4", "v5", "v6"], "name": "Part A "}], "name": "Peter's copy of Factorising Quadratic Equations with $x^2$ Coefficients of 1", "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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\nThe first has two negative roots, the second has one negative and one positive, and the third is the difference of two squares.
"}, "ungrouped_variables": [], "statement": "Factorise the following quadratic equations.
\n", "variablesTest": {"maxRuns": 100, "condition": ""}, "rulesets": {}, "tags": ["Factorisation", "factorisation", "Factorising quadratic equations", "factorising quadratic equations", "taxonomy"], "type": "question", "contributors": [{"name": "Peter Knowles", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1802/"}]}]}], "contributors": [{"name": "Peter Knowles", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1802/"}]}