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Consider the quadratic $\\simplify[all,expandBrackets]{({a}*x + {b})*(x+{c})}$.

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What are the the two linear factors of this quadratic?

\n

[[0]] and [[1]].

You can use the quadratic formula to deduce that $\\simplify[all,expandBrackets]{({a}*x + {b})*(x+{c})}$ has roots:

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$x = \\frac{\\simplify{-({a}*{c}+{b})}\\pm\\sqrt{ (\\var{a*c+b})^2 - 4\\times(\\var{a})\\times(\\var{b*c}) }}{2\\times \\var{a}} = \\var{-1*c} \\text{ or } \\displaystyle \\simplify{-1*{b}/{a}}.$

\n

The 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:

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$C \\times (x + \\simplify{{b}/{a}}) \\times (x - \\var{-1*c})$

\n

for some constant term $C$. The only thing left to do is determine the value of the constant which makes:

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$C \\times (x + \\simplify{{b}/{a}}) \\times (x - \\var{-1*c}) = \\simplify[all,expandBrackets]{({a}*x + {b})*(x+{c})}$.

\n

Equating 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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a

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Quadratic factorisation that does not rely upon pattern matching.

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