a) Áp dụng định lý Vi-ét, ta có:
\(\left\{ \matrix{ {x_1} + {x_2} = - {b \over a} \hfill \cr {x_1}.{x_2} = {c \over a} \hfill \cr} \right.\)
Do đó:
\(\eqalign{
& a{x^2} + {\rm{ }}bx + c = 0 = a({x^2} + {b \over a}x + {c \over a}) \cr&= a{\rm{[}}{{{x}}^2} - ({x_1} + {x_2})x + {x_1}{x_2}{\rm{]}} \cr
& = a{\rm{[x(x}}\,{\rm{ - }}\,{{\rm{x}}_1}) - {x_2}(x\, - \,{x_1}){\rm{]}} = a(x - {x_1})(x - {x_2}) \cr} \)
b) Ta có:
\(f(x) = - 2{x^2} - 7x + 4 = 0 \Leftrightarrow \left[ \matrix{ x = - 4 \hfill \cr x = {1 \over 2} \hfill \cr} \right.\)
Do đó: \(f(x) = - 2(x + 4)(x - {1 \over 2}) = (x + 4)(1 - 2x)\)
Ta có:
\(\eqalign{
& g(x) = (\sqrt 2 + 1){x^2} - 2(\sqrt 2 + 1)x + 2 = 0 \cr
& \Leftrightarrow \left[ \matrix{
x = \sqrt 2 \hfill \cr
x = {{\sqrt 2 } \over {\sqrt 2 + 1}} \hfill \cr} \right. \cr} \)
Do đó: \(g(x) = (\sqrt 2 + 1)(x - \sqrt 2 )(x - {{\sqrt 2 } \over {\sqrt 2 + 1}}) \)
\(= (x - \sqrt 2 ){\rm{[}}(\sqrt 2 + 1)x\, - \sqrt 2 {\rm{]}}\)