Đáp án:
\({\left( {x - 1} \right)^2}\left( { - 4{x^2} - 8x - 5} \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
(8x - 4{x^2} - 1)({x^2} + 2x + 1) - 4({x^2} + x + 1)\\
= 8{x^3} + 16{x^2} + 8x - 4{x^4} - 8{x^3} - 4{x^2} - {x^2} - 2x - 1 - 4{x^2} - 4x - 4\\
= - 4{x^4} + 7{x^2} + 2x - 5\\
= - 4{x^4} + 4{x^3} - 4{x^3} + 4{x^2} + 3{x^2} - 3x + 5x - 5\\
= - 4{x^3}\left( {x - 1} \right) - 4{x^2}\left( {x - 1} \right) + 3x\left( {x - 1} \right) + 5\left( {x - 1} \right)\\
= \left( {x - 1} \right)\left( { - 4{x^3} - 4{x^2} + 3x + 5} \right)\\
= \left( {x - 1} \right)\left( { - 4{x^3} + 4{x^2} - 8{x^2} + 8x - 5x + 5} \right)\\
= \left( {x - 1} \right)\left[ { - 4{x^2}\left( {x - 1} \right) - 8x\left( {x - 1} \right) - 5\left( {x - 1} \right)} \right]\\
= {\left( {x - 1} \right)^2}\left( { - 4{x^2} - 8x - 5} \right)
\end{array}\)
