Question

$2x^5+x^4-2x-1=0\quad$

solve

Answer 1

$2x^5+x^4-2x-1=0\quad :\quad x=-\frac{1}{2},\:x=i,\:x=-i,\:x=-1,\:x=1$

$\mathrm{Solve\:by\:factoring}$

$\mathrm{Factor\:}2x^5+x^4-2x-1:\quad \left(2x+1\right)\left(x^2+1\right)\left(x+1\right)\left(x-1\right)$

$\left(2x+1\right)\left(x^2+1\right)\left(x+1\right)\left(x-1\right)=0$

$\mathrm{Using\:the\:Zero\:Factor\:Principle:\quad \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

$\mathrm{Solve\:}\:2x+1=0:\quad x=-\frac{1}{2}$

$\mathrm{Solve\:}\:x^2+1=0:\quad x=i,\:x=-i$

$\mathrm{Solve\:}\:x+1=0:\quad x=-1$

$\mathrm{Solve\:}\:x-1=0:\quad x=1$

$\mathrm{The\:solutions\:are}$

$x=-\frac{1}{2},\:x=i,\:x=-i,\:x=-1,\:x=1$

Answer 2

Explanation:

$\sf \to 2x^5+x^4-2x-1=0 \\ \\ \: \sf \to \: x( {x}^{4} + {x}^{3} - 2) - 1 = 0 \\ \\ \: \sf \to \: \: (x - 1)({x}^{4} + {x}^{3} - 2) = 0 \\ \\ : \sf \to \: (x - 1) \: {x}^{2} ( {x}^{2} + x) - 2 = 0 \\ \\ \sf \to \red{(x - 1) \: ( {x}^{2} - 2)( {x}^{2} + x)}$