Quadratic factorisation that does not rely upon pattern matching.
\nOne could also use the pattern matching syntax to check automatically; this is programatically harder.
", "licence": "Creative Commons Attribution-ShareAlike 4.0 International"}, "statement": "Consider the quadratic $ \\simplify[all,expandBrackets]{({a}*x + {b})*(x+{c})}$.
", "advice": "You can use the quadratic formula to deduce that $\\simplify[all,expandBrackets]{({a}*x + {b})*(x+{c})}$ has roots:
\n$ x = \\frac{\\simplify{-({a}*{c}+{b})}\\pm\\sqrt{ (\\var{a*c+b})^2 - 4\\times(\\var{a})\\times(\\var{b*c}) }}{2\\times \\var{a}} = \\var{-1*c} \\text{ or } \\displaystyle \\simplify{-1*{b}/{a}}.$
\nThe roots determine the factors, but only upto a constant. In general, a quadratic with roots $ \\var{-1*c}$ and $\\simplify{-1*{b}/{a}}$ has the form:
\n$C \\times (x + \\simplify{{b}/{a}}) \\times (x - \\var{-1*c})$
\nfor some constant term $C$. The only thing left to do is determine the value of the constant which makes:
\n$C \\times (x + \\simplify{{b}/{a}}) \\times (x - \\var{-1*c}) = \\simplify[all,expandBrackets]{({a}*x + {b})*(x+{c})}$.
\nEquating the coefficients of the $x^2$ terms in the left and right hand sides shows that $C=\\var{a}$. So
\n$ (\\var{a}x + \\var{b}) \\times (x-\\var{-1*c}) = \\simplify[all,expandBrackets]{({a}*x + {b})*(x+{c})}$.
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\n[[0]] and [[1]].
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