The number of distinct real roots of the equation x5(x3−x2−x+1)+x(3x3−4x2−2x+4)−1=0 is \_\_\_\_\_\_\_\_\_\_\_\_.
Answer: 3
Solutions · 1
Official solution
Expanding: x8−x7−x6+x5+3x4−4x3−2x2+4x−1=0.
Grouping terms:
\[x^7(x-1)-x^5(x-1)+3x^3(x-1)-x(x^2-1)+2x(1-x)+(x-1)=0\]
\[\implies (x-1)\big(x^7-x^5+3x^3-x^2-3x+1\big)=0\]
\[\implies (x-1)(x^2-1)\big(x^5+3x-1\big)=0.\]
So x=±1 are roots (2 real roots), and x5+3x−1 is strictly increasing (derivative 5x4+3>0 always), so it has exactly 1 real root. Total: 3 distinct real roots.