Question

Solve for x and please show steps!
(2x+1)(x-1)^2+(x+5)(2x+1)^2=0

Answer 1

\[(2x+1)(x^2-2x+1)+(x+5)(4x^2+4x+1)=0\]
2x^3-4x^2+2x+x^2-2x+1+4x^3+4x^2+x+20x^2+20x+5=0

Answer 2

yes i got that part then i got 6x^3+21x^2+21x+6
and i dont know what to do after that

Answer 3

\[6x^3+29x^2+21x+6=0\]

Answer 4

The possible rational roots are:
\[\pm1,\pm2,\pm3,\pm6,\pm\frac{1}{6}, \pm\frac{1}{3},+\frac{1}{2},\pm\frac{2}{3},\pm\frac{3}{2}\]

Answer 5

Descartes rule of signs tells us there are no positive real roots so let's use ssynthetic division and try to find some negative roots.