Answer:
The polynomial which is factored completely is \(2g^2+5g+4\) as these are not further factored.
Step-by-step explanation:
To find : Which polynomial is factored completely?
Solution :
We factor the given polynomial one by one,
1) \(g^5-g\)
\(g^5-g=g(g^4-1)\)
\(g^5-g=g(g^2-1)(g^2+1)\)
\(g^5-g=g(g-1)(g+1)(g^2+1)\)
2) \(4g^3 + 18g^2+20g\)
\(4g^3 + 18g^2+20g=2g(2g^{2}+9g+10)\)
\(4g^3 + 18g^2+20g=2g(2g^{2}+5g+4g+10)\)
\(4g^3 + 18g^2+20g=2g[g(2g+5)+2(2g+5)]\)
\(4g^3 + 18g^2+20g=2g(2g+5)(g+2)\)
3) \(24g^2 - 6g^4\)
\(24g^2 - 6g^4=6g^{2}(4-g^{2})\)
\(24g^2 - 6g^4=6g^{2}(2-g)(2+g)\)
4) \(2g^2+5g+4\)
This expression is can not be factored with rational number as
\(D=b^2-4ac\\D=5^2-4(2)(4)\\D=25-32\\D=-7\)
Discriminant D<0 so there is no rational roots.
So, The polynomial which is factored completely is \(2g^2+5g+4\) as these are not further factored.
