Let's go through each polynomial and check if it is factored completely.
1. [tex]\( g^5 - 9 \)[/tex]:
- This polynomial can be written as [tex]\( g^5 - 3^2 \)[/tex].
- However, this isn't a straightforward factorization using common algebraic identities directly.
- We can see that it cannot be factored into simpler polynomials with integer coefficients.
2. [tex]\( 4g^3 + 18g^2 + 20g \)[/tex]:
- First, identify any common factors in all the terms. In this case, each term has a factor of [tex]\(2g\)[/tex].
- Factoring out [tex]\(2g\)[/tex]:
[tex]\[
4g^3 + 18g^2 + 20g = 2g(2g^2 + 9g + 10).
\][/tex]
- Next, factor the quadratic polynomial [tex]\(2g^2 + 9g + 10\)[/tex].
- By finding the roots or using factorization techniques, we discover:
[tex]\[
2g^2 + 9g + 10 = (g + 2)(2g + 5).
\][/tex]
- Thus, the fully factored form is:
[tex]\[
4g^3 + 18g^2 + 20g = 2g(g + 2)(2g + 5).
\][/tex]
3. [tex]\( 24g^2 - 6g^4 \)[/tex]:
- First, observe the common factors. Both terms have a common factor of [tex]\(6g^2\)[/tex].
- Factoring out [tex]\(6g^2\)[/tex]:
[tex]\[
24g^2 - 6g^4 = 6g^2(4 - g^2).
\][/tex]
- Notice that [tex]\(4 - g^2\)[/tex] is a difference of squares and can be further factored:
[tex]\[
4 - g^2 = (2 - g)(2 + g).
\][/tex]
- Therefore, the complete factorization is:
[tex]\[
24g^2 - 6g^4 = 6g^2(2 - g)(2 + g).
\][/tex]
4. [tex]\( 2g^2 + 5g + 4 \)[/tex]:
- Checking if this quadratic can be factored.
- We look for two numbers that multiply to [tex]\(2 \cdot 4 = 8\)[/tex] and add to [tex]\(5\)[/tex].
- There are no such integer pairs.
- Therefore, it cannot be factored further and remains as:
[tex]\[
2g^2 + 5g + 4.
\][/tex]
Among the given polynomials, the polynomial [tex]\(4g^3 + 18g^2 + 20g\)[/tex] which factors to [tex]\(2g(g + 2)(2g + 5)\)[/tex] is the one that is factored completely.
