Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
A = {x^3} - 30{x^2} - 31x + 1\\
= \left( {{x^3} - 31{x^2}} \right) + \left( {{x^2} - 31x} \right) + 1\\
= {x^2}\left( {x - 31} \right) + x\left( {x - 31} \right) + 1\\
= \left( {x - 31} \right)\left( {{x^2} + x} \right) + 1\\
x = 31 \Rightarrow \left( {x - 31} \right)\left( {{x^2} + x} \right) = 0 \Rightarrow A = 1\\
b,\\
B = {x^5} - 15{x^4} + 16{x^3} - 29{x^2} + 13x\\
= \left( {{x^5} - 14{x^4}} \right) - \left( {{x^4} - 14{x^3}} \right) + \left( {2{x^3} - 28{x^2}} \right) - \left( {{x^2} - 14x} \right) - x\\
= {x^4}\left( {x - 14} \right) - {x^3}\left( {x - 14} \right) + 2{x^2}\left( {x - 14} \right) - x\left( {x - 14} \right) - x\\
= \left( {x - 14} \right).\left( {{x^4} - {x^3} + 2{x^2} - x} \right) - x\\
x = 14 \Rightarrow \left( {x - 14} \right).\left( {{x^4} - {x^3} + 2{x^2} - x} \right) = 0\\
\Rightarrow B = - x = - 14\\
c,\\
C = {x^{14}} - 10{x^{13}} + 10{x^{12}} - 10{x^{11}} + .... + 10{x^2} - 10x + 10\\
= \left( {{x^{14}} - 9{x^{13}}} \right) - \left( {{x^{13}} - 9{x^{12}}} \right) + \left( {{x^{12}} - 9{x^{11}}} \right) - ...... + \left( {{x^2} - 9x} \right) - \left( {x - 9} \right) + 1\\
= {x^{13}}\left( {x - 9} \right) - {x^{12}}\left( {x - 9} \right) + {x^{11}}\left( {x - 9} \right) - ..... + x\left( {x - 9} \right) - \left( {x - 9} \right) + 1\\
= \left( {x - 9} \right).\left( {{x^{13}} - {x^{12}} + {x^{11}} - ..... + x - 1} \right) + 1\\
x = 9 \Rightarrow C = 1
\end{array}\)
