// Numbas version: exam_results_page_options {"name": "Solving a monic quadratic by factorising", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"functions": {}, "ungrouped_variables": ["a", "b", "linear", "const", "monicsoln"], "name": "Solving a monic quadratic by factorising", "tags": ["binomial", "factorisation", "Factorisation", "factorising", "factors", "Factors", "monic", "quadratic", "quadratics", "solving"], "preamble": {"css": "", "js": ""}, "advice": "", "rulesets": {}, "parts": [{"stepsPenalty": "1", "prompt": "

 Solve the following quadratic by factorisation:

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$\\simplify{x^2+{linear}x+{const}}$$=$0
$\\Longrightarrow$ [[0]]$=$0
$\\Longrightarrow$$x$$=$[[1]]
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Note: In the first gap, enter the quadratic in factored form.

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Note: In the second gap, if $x=1,2$, enter set(1,2).

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Ensure you factorise the expression.

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Ensure you factorise the expression.

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Since $(x+a)(x+b)=x^2+(a+b)x+ab$, when we are factorising a quadratic, such as $x^2+cx+d$, we must find the numbers $a$ and $b$ such that $c=a+b$ and $d=ab$.

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In the case of $\\simplify{x^2+{linear}x+{const}}$ we ask

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what two numbers add to give $\\var{linear}$ and multiply to give $\\var{const}$? 

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Therefore the numbers must be $\\var{a}$ and $\\var{b}$, that is 

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$\\simplify{x^2+{linear}x+{const}}=(\\simplify{x+{a}})(\\simplify{x+{b}}).$

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You can check this by expanding the binomial product.

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Now, using the null factor law we have, either $\\simplify{x+{a}}=0$ or $\\simplify{x+{b}}=0$. In otherwords, either $x=\\var{-a}$ or $\\var{-b}$.

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\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$\\simplify{x^2+{linear}x+{const}}$$=$0
$\\Longrightarrow$$ (\\simplify{x+{a}})(\\simplify{x+{b}})$$=$0
$\\Longrightarrow$$x$$=$$\\var{-a},\\var{-b}$
\n

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