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Factorise three quadratic equations of the form $x^2+bx+c$.

The first has two negative roots, the second has one negative and one positive, and the third is the difference of two squares.

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Quadratic expressions of the form \\[x^2+bx+c\\] can be factorised to create an equation of the form \\[(x+m)(x+n)\\text{.}\\]

Note also that \\[(x+m)(x+n)=x^2+(m+n)x+(m \\times n)\\text{.}\\]

Thus to factorise an expression of the form $x^2+bx+c$, we need to find two numbers which add together to make $b$, and multiply together to make $c$.

(a)

\\[\\simplify{x^2+{v1+v2}x+{v1*v2}}\\]

We need to find two values that add together to make $\\var{v1+v2}$ and multiply together to make $\\var{v1*v2}$.

\\[\\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 expression is

\\[\\simplify{(x+{v1})(x+{v2})}\\text{.}\\]

(b)

We need to find two numbers that multiply to give $\\var{v3*v4}$ and add together to give $\\var{v3+v4}$.

\\[\\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 expressionis

\\[\\simplify{(x+{v3})(x+{v4})}\\text{.}\\]

Hence to solve $\\simplify{x^2+{v3+v4}x+{v3*v4}}=0$ we have:

$\\simplify{(x+{v3})(x+{v4})}=0$

So either $\\simplify{(x+{v3})}=0$ or $\\simplify{(x+{v4})}=0$

which gives solutions of $x=\\var{-v3}$ or $x=\\var{-v4}$

(c)

When factorising the quadratic expression

\\[\\simplify{x^2+{v5*v6}}\\]

we need to find two values that add together to make $0$ and multiply together to make $\\var{v5*v6}$.

\\begin{align}
\\var{v5} \\times \\var{v6}& = \\var{v5*v6}\\\\
\\simplify[]{ {v5} + {v6}} &= 0 \\\\
\\end{align}

So the factorised form of the expression is

\\[\\simplify{(x+{v5})(x+{v6})}\\text{.}\\]

Note this is a common form of quadratic called a \"Difference of Two Squares\". In general $x^2-a^2=(x-a)(x+a)$

", "variables": {"v3": {"templateType": "anything", "name": "v3", "description": "", "group": "Part A ", "definition": "random(-8..-1)"}, "v2": {"templateType": "anything", "name": "v2", "description": "", "group": "Part A ", "definition": "random(2..6 except v1)"}, "v6": {"templateType": "anything", "name": "v6", "description": "", "group": "Part A ", "definition": "-v5"}, "v1": {"templateType": "anything", "name": "v1", "description": "", "group": "Part A ", "definition": "random(1..10)"}, "v5": {"templateType": "anything", "name": "v5", "description": "", "group": "Part A ", "definition": "random(2..10)"}, "v4": {"templateType": "anything", "name": "v4", "description": "", "group": "Part A ", "definition": "random(1..10 except -v3)"}}, "preamble": {"css": "", "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}"}, "ungrouped_variables": [], "parts": [{"variableReplacements": [], "customMarkingAlgorithm": "", "unitTests": [], "gaps": [{"variableReplacements": [], "customMarkingAlgorithm": "", "unitTests": [], "vsetRangePoints": 5, "answer": "(x+{v1})(x+{v2})", "type": "jme", "failureRate": 1, "showPreview": true, "checkingAccuracy": 0.001, "scripts": {"mark": {"script": "question.is_factorised(this);", "order": "after"}}, "expectedVariableNames": [], "marks": 1, "extendBaseMarkingAlgorithm": true, "checkingType": "absdiff", "showFeedbackIcon": true, "checkVariableNames": false, "variableReplacementStrategy": "originalfirst", "vsetRange": [0, 1], "showCorrectAnswer": true}], "type": "gapfill", "prompt": "

$\\simplify{x^2+{v1+v2}x+{v1*v2}}$

[[0]]

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$\\simplify{x^2+{v3+v4}x+{v3*v4}}$

[[0]]

Hence solve $\\simplify{x^2+{v3+v4}x+{v3*v4}}=0$ (Put your answers in ascending order!)

$x=$[[1]] or $x=$[[2]]

$\\simplify{x^2+{v5*v6}}$

[[0]]

Factorise the following quadratic expressions.

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