Đáp án:
\(P = \left( {x + 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)\left( {2x + 1} \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
P = 2{x^4} - 7{x^3} - 2{x^2} + 13x + 6\\
a)\,\,\\
P = 2{x^4} - 7{x^3} - 2{x^2} + 13x + 6\\
P = 2{x^4} + 2{x^3} - 9{x^3} - 9{x^2} + 7{x^2} + 7x + 6x + 6\\
P = 2{x^3}\left( {x + 1} \right) - 9{x^2}\left( {x + 1} \right) + 7x\left( {x + 1} \right) + 6\left( {x + 1} \right)\\
P = \left( {x + 1} \right)\left( {2{x^3} - 9{x^2} + 7x + 6} \right)\\
P = \left( {x + 1} \right)\left[ {2{x^3} - 4{x^2} - 5{x^2} + 10x - 3x + 6} \right]\\
P = \left( {x + 1} \right)\left[ {2{x^2}\left( {x - 2} \right) - 5x\left( {x - 2} \right) - 3\left( {x - 2} \right)} \right]\\
P = \left( {x + 1} \right)\left( {x - 2} \right)\left( {2{x^2} - 5x - 3} \right)\\
P = \left( {x + 1} \right)\left( {x - 2} \right)\left( {2{x^2} - 6x + x - 3} \right)\\
P = \left( {x + 1} \right)\left( {x - 2} \right)\left[ {2x\left( {x - 3} \right) + \left( {x - 3} \right)} \right]\\
P = \left( {x + 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)\left( {2x + 1} \right)\\
b)\,\,\\
x - 2,\,\,x - 3\,\,la\,\,hai\,\,nguyen\,so\,\,lien\,\,tiep \Rightarrow \left( {x - 2} \right)\left( {x - 3} \right)\,\, \vdots \,\,2\\
\Rightarrow P\,\, \vdots \,\,2\\
Chung\,\,minh\,\,P\,\, \vdots \,\,3\\
TH1:\,\,x = 3k\,\,\left( {k \in Z} \right) \Rightarrow x - 3\,\, \vdots \,\,3\\
\Rightarrow P\,\, \vdots \,\,3\\
TH2:\,\,x = 3k + 1\,\,\left( {k \in Z} \right) \Rightarrow 2x + 1 = 6k + 3\,\, \vdots \,\,3\\
\Rightarrow P\,\, \vdots \,\,3\\
TH3:\,\,x = 3k + 2\,\,\left( {k \in Z} \right) \Rightarrow x + 1 = 3k + 3\,\, \vdots \,\,3\\
\Rightarrow P\,\, \vdots \,\,3\\
Do\,\,do\,\,P\,\, \vdots \,\,3\,\,\forall x \in Z\\
Vay\,\,P\,\, \vdots \,\,6.
\end{array}\)
