which polynomial is factored completely?\n$g^{5}-g$\n$4g^{3}+18g^{2}+20g$\n$24g^{2}-6g^{4}$\n$2g^{2}+5g + 4$

Explanation:

Step1: Factorize $g^{5}-g$

$g^{5}-g = g(g^{4}-1)=g(g^{2} + 1)(g^{2}-1)=g(g^{2}+1)(g + 1)(g - 1)$

Step2: Factorize $4g^{3}+18g^{2}+20g$

$4g^{3}+18g^{2}+20g=2g(2g^{2}+9g + 10)=2g(2g^{2}+4g+5g + 10)=2g[2g(g + 2)+5(g + 2)]=2g(2g + 5)(g+2)$

Step3: Factorize $24g^{2}-6g^{4}$

$24g^{2}-6g^{4}=6g^{2}(4 - g^{2})=6g^{2}(2 + g)(2 - g)$

Step4: Analyze $2g^{2}+5g + 4$

For the quadratic polynomial $ax^{2}+bx + c$ ($a = 2$, $b = 5$, $c = 4$), the discriminant $\Delta=b^{2}-4ac=5^{2}-4\times2\times4=25 - 32=-7<0$. So it cannot be factored over the real - numbers.

Answer:

$2g^{2}+5g + 4$