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Factorise the following into linear factors. That is, write the quadratic as a product of terms that look like $ax+b$ where $a$ and $b$ are real numbers.
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\n\n\n\n", "stepsPenalty": "1", "steps": [{"type": "information", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "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 $d=ab$.
\n\nIn the case of $\\simplify{x^2+{linear}x+{const}}$ we ask
\nwhat two numbers add to give $\\var{linear}$ and multiply to give $\\var{const}$?
\nTherefore the numbers must be $\\var{a}$ and $\\var{b}$, that is
\n$\\simplify{x^2+{linear}x+{const}}=(\\simplify{x+{a}})(\\simplify{x+{b}}).$
\nYou can check this by expanding the binomial product.
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", "showStrings": false, "strings": ["(", ")"], "partialCredit": 0}}], "steps": [{"prompt": "Since $\\simplify{{aa}x^2}$ is $\\simplify{{a}x}$ squared and $\\var{bb}$ is $\\var{b}$ squared, we can recognise $\\simplify{{aa}x^2-{bb}}$ as a difference of two squares.
\nRecalling that $(a+b)(a-b)=a^2-b^2$, we have
\n$\\simplify{{aa}x^2-{bb}}=(\\simplify{{a}x+{b}})(\\simplify{{a}x-{b}})$
\nNotice there is a common factor of $\\var{gg}$ that we can deal with first
\n$\\simplify{{aa}x^2-{bb}}=\\simplify{{gg}({aa/gg}x^2-{bb/gg})}.$
\nNext, notice the remaining expression is a difference of two squares. Recalling that $(a+b)(a-b)=a^2-b^2$, we have
\n$\\simplify{{aa}x^2-{bb}}=\\simplify{{gg}({aa/gg}x^2-{bb/gg})}=\\simplify{({gg})({a/g}x+{b/g})({a/g}x-{b/g})}$
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", "showStrings": false, "strings": ["(", ")"], "partialCredit": 0}}], "steps": [{"prompt": "We should recognise this as a difference of two squares, where $\\simplify{{c[0]}^2/{c[2]}^2 x^2}$ is $\\left(\\simplify{{c[0]}/{c[2]}x}\\right)^2$ and $\\simplify{{c[1]}^2/{c[3]}^2}$ is $\\left(\\simplify{{c[1]}/{c[3]}}\\right)^2$. Therefore
\n$\\simplify{{c[0]}^2/{c[2]}^2 x^2-{c[1]}^2/{c[3]}^2}=\\simplify{({c[0]}/{c[2]}x+{c[1]}/{c[3]})({c[0]}/{c[2]}x-{c[1]}/{c[3]})}.$
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\nIn fact, $\\simplify{{aa}x^2+{mid}x+{bb}}$ is also a perfect square, since
\nThat is, $\\simplify{{aa}x^2+{mid}x+{bb}}=\\simplify{({a}x+{b})^2}$.
\nEach bracket has a common factor of $\\var{g}$, so we call move both of them to the front, to get
\n$\\simplify{{aa}x^2+{mid}x+{bb}}=\\simplify{{gg}({a/g}x+{b/g})^2}$.
\nYou can always check your factorisation by expanding.
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\n", "stepsPenalty": "1", "steps": [{"type": "information", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "Recall that $(a+b)^2=(a+b)(a+b)=a^2+2ab+b^2$ is called a perfect square.
\nIn fact, $\\simplify{{c[0]^2}/{c[1]^2} x^2+{2c[0]*d}/{(c[1]*c[2])}x+{d^2}/{c[2]^2}}$ is also a perfect square, since
\nThat is,
\n$\\simplify{{c[0]^2}/{c[1]^2} x^2+{2c[0]*d}/{(c[1]*c[2])}x+{d^2}/{c[2]^2}}=\\simplify{({c[0]}/{c[1]}x+{d}/{c[2]})^2}$.
\n\nYou can always check your factorisation by expanding.
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\n", "stepsPenalty": "1", "steps": [{"type": "information", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "There are a few different ways to do the working for these questions, here is one method that uses factorisation by grouping.
\n
Given $\\simplify{{c[1]}x^2+{d[1]+b[1]*c[1]}x+{b[1]*d[1]}}$, we
As always, you can check your factorisation by expanding.
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\n", "stepsPenalty": "1", "steps": [{"type": "information", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "There are a few different ways to do the working for these questions, here is one method that uses factorisation by grouping.
\n
Given $\\simplify{{a[0]*c[0]}x^2+{a[0]*d[0]+b[0]*c[0]}x+{b[0]*d[0]}}$, we
As always, you can check your factorisation by expanding.
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