Đáp án:
c) \(\left( {x - 2} \right)\left( {{x^2} + 2x + 2} \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a){x^3} - 4{x^2} - 8x + 8\\
= {x^3} + 2{x^2} - 6{x^2} - 12x + 4x + 8\\
= {x^2}\left( {x + 2} \right) - 6x\left( {x + 2} \right) + 4\left( {x + 2} \right)\\
= \left( {x + 2} \right)\left( {{x^2} - 6x + 4} \right)\\
b){x^2} - 5x - 14\\
= {x^2} - 7x + 2x - 14\\
= x\left( {x - 7} \right) + 2\left( {x - 7} \right)\\
= \left( {x - 7} \right)\left( {x + 2} \right)\\
c){x^3} - 2x - 4\\
= {x^3} - 2{x^2} + 2{x^2} - 4x + 2x - 4\\
= {x^2}\left( {x - 2} \right) + 2x\left( {x - 2} \right) + 2\left( {x - 4} \right)\\
= \left( {x - 2} \right)\left( {{x^2} + 2x + 2} \right)
\end{array}\)
\(\begin{array}{l}
B3:\\
a)12x + 7 = 0\\
\to x = - \dfrac{7}{{12}}\\
b)15x - 5 = 0\\
\to x = \dfrac{1}{3}\\
c)16x - 6x = 2\\
\to 10x = 2\\
\to x = \dfrac{1}{5}\\
d)3x + 5x = - 7\\
\to 8x = - 7\\
\to x = - \dfrac{7}{8}
\end{array}\)
