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Factorising a quadratic expression of the form $ax^2+bx+c$, where $a>1$.

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Factorise the following quadratic expression:

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\\[ \\simplify[unitFactor]{{a}x^2+{b}x+{c}} \\]

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For  some quadratic expressions of the form \\[ ax^2+bx+c,\\] where $a \\neq 1$, we are able to factorise the expression into the form \\[(mx+p)(nx+q),\\] for integers $m$, $n$, $p$, and $q$.

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This is more difficult than when $a=1$, but there is a process which allows us to factorise certain quadratics of this type. 

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The first step is to identify two numbers which multiply to give $ac$ and add to give $b$.

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So, for $\\simplify[unitFactor]{{a}x^2+{b}x+{c}}$, we have

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\\[ a \\times c = \\var{a*c} \\quad\\text{and}\\quad b=\\var{b}.\\] 

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In this case, the numbers we need are $\\var{m*q}$ and $\\var{n*p}$.

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Next, we want to rewrite our quadratic expression so that we have 2 linear terms with these coefficients:

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\\[\\simplify[unitFactor,!canonicalOrder]{{a}x^2+{m*q}x+{n*p}x+{c}}.\\]

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We are now able to factor the first two terms and the last two terms:

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\\[\\simplify[unitFactor,!canonicalOrder]{{m}x({n}x+{q})+{p}({n}x+{q})}.\\]

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We can see that we now have a common factor of $(\\simplify{{n}x+{q}})$, so rewriting our expression in terms of this factor, we get:

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\\[\\simplify[unitFactor]{({n}x+{q})({m}x+{p})},\\]

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as required.

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