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Given two quadratic expressions $f(x)$ and $g(x)$, calculate $f(x)(g(x)-f(x))$. 

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Given two functions, \\[ f(x) = \\simplify{{a1}x^2+{b1}x+{c1}}, \\quad g(x)=\\simplify{{a2}x^2+{b2}x+{c2}},\\]

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calculate \\[ f(x)\\left[g(x)-f(x)\\right].\\]

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To calculate $f(x)\\left[g(x)-f(x)\\right]$, we want to first find $g(x)-f(x)$:

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\\[ \\begin{split} g(x)-f(x) &\\,= \\left(\\simplify{{a2}x^2+{b2}x+{c2}}\\right) - \\left(\\simplify{{a1}x^2+{b1}x+{c1}}\\right) \\\\ &\\,= \\simplify[all,!collectNumbers,!cancelTerms]{{a2}x^2+{b2}x+{c2}-{a1}x^2-{b1}x - {c1}} \\\\&\\,= \\simplify[all]{{a2-a1}x^2+{b2-b1}x+{c2-c1}}. \\end{split} \\]

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We now want to multiply this result by $f(x)$:

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\\[ \\begin{split} f(x) \\left(\\simplify{{a2-a1}x^2+{b2-b1}x+{c2-c1}}\\right) &\\,=  \\left(\\simplify{{a1}x^2+{b1}x+{c1}}\\right)\\left(\\simplify{{a2-a1}x^2+{b2-b1}x+{c2-c1}}\\right) \\\\ &\\,= \\simplify{{a1}x^2({a2-a1}x^2+{b2-b1}x+{c2-c1})+{b1}x({a2-a1}x^2+{b2-b1}x+{c2-c1})+{c1}({a2-a1}x^2+{b2-b1}x+{c2-c1})}. \\end{split} \\]

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We can now expand each of the brackets and collect similar terms:

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\\[ \\begin{split} f(x)\\left[g(x)-f(x)\\right] &\\,= \\simplify{{a1}x^2({a2-a1}x^2+{b2-b1}x+{c2-c1})+{b1}x({a2-a1}x^2+{b2-b1}x+{c2-c1})+{c1}({a2-a1}x^2+{b2-b1}x+{c2-c1})} \\\\ &\\,= \\simplify[all, !cancelTerms, !collectNumbers]{{a1*(a2-a1)}x^4+{a1*(b2-b1)}x^3+{a1*(c2-c1)}x^2 + {b1*(a2-a1)}x^3+{b1*(b2-b1)}x^2+{b1*(c2-c1)}x + {c1*(a2-a1)}x^2+{c1*(b2-b1)}x+{c1*(c2-c1)}} \\\\ &\\,= \\simplify{{a1*da}x^4+{a1*db+da*b1}x^3+{a1*dc+da*c1+b1*db}x^2+{b1*dc+db*c1}x+{c1*dc}} \\end{split} \\]

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