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

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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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$\\simplify{x^2+{v3+v4}x+{v3*v4}}=0$

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[[0]] $=0$

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", "nameToCompare": "", "partialCredit": 0, "pattern": "(+-x^$n? + +-$n)* * $z"}, "answer": "(x+{v5})(x+{v6})", "variableReplacements": []}], "marks": 0, "sortAnswers": false, "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "customMarkingAlgorithm": "", "scripts": {}, "prompt": "$\\simplify{x^2+{v5*v6}}=0$\n [[0]]$=0$", "customName": "", "unitTests": []}], "name": "what the heck?", "advice": " Quadratic equations of the form \n \$x^2+bx+c=0\$ \n can be factorised to create an equation of the form \n \$(x+m)(x+n)=0\\text{.}\$ \n When we expand a factorised quadratic expression we obtain \n \$(x+m)(x+n)=x^2+(m+n)x+(m \\times n)\\text{.}\$ \n To factorise an equation of the form$x^2+bx+c$, we need to find two numbers which add together to make$b$, and multiply together to make$c$. \n #### a) \n \$\\simplify{x^2+{v1+v2}x+{v1*v2}=0}\$ \n We need to find two values that add together to make$\\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} \ \n So the factorised form of the equation is \n \$\\simplify{(x+{v1})(x+{v2})}=0\\text{.}\$ \n \n #### b) \n We 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} \ \n So the factorised form of the equation is \n \$\\simplify{(x+{v3})(x+{v4})}=0\\text{.}\$ \n #### c) \n When factorising the quadratic expression \n \$\\simplify{x^2+{v5*v6}=0}\$ \n we need to find two values that add together to make0$and multiply together to make$\\var{v5*v6}\$.

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

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So the factorised form of the equation is

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\$\\simplify{(x+{v5})(x+{v6})}=0\\text{.}\$

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