To determine which of the given choices are factors of the polynomial [tex]\(3x^3 + 18x^2 + 27x\)[/tex], we need to test each factor by dividing the polynomial by each choice and checking if the result is a polynomial with no remainder.
1. Testing [tex]\(9x\)[/tex]:
Dividing [tex]\(3x^3 + 18x^2 + 27x\)[/tex] by [tex]\(9x\)[/tex]:
[tex]\[
\frac{3x^3 + 18x^2 + 27x}{9x} = \frac{3x^3}{9x} + \frac{18x^2}{9x} + \frac{27x}{9x} = \frac{3}{9} x^2 + \frac{18}{9} x + \frac{27}{9} = \frac{1}{3} x^2 + 2x + 3
\][/tex]
The result, [tex]\(\frac{1}{3} x^2 + 2x + 3\)[/tex], is a polynomial. Thus, [tex]\(9x\)[/tex] is a factor of [tex]\(3x^3 + 18x^2 + 27x\)[/tex].
2. Testing [tex]\(x^3\)[/tex]:
Dividing [tex]\(3x^3 + 18x^2 + 27x\)[/tex] by [tex]\(x^3\)[/tex]:
[tex]\[
\frac{3x^3 + 18x^2 + 27x}{x^3} = \frac{3x^3}{x^3} + \frac{18x^2}{x^3} + \frac{27x}{x^3} = 3 + \frac{18}{x} + \frac{27}{x^2}
\][/tex]
The result, [tex]\(3 + \frac{18}{x} + \frac{27}{x^2}\)[/tex], is not a polynomial because it contains negative powers of [tex]\(x\)[/tex]. Thus, [tex]\(x^3\)[/tex] is not a factor of [tex]\(3x^3 + 18x^2 + 27x\)[/tex].
3. Testing [tex]\(x+3\)[/tex]:
Dividing [tex]\(3x^3 + 18x^2 + 27x\)[/tex] by [tex]\(x+3\)[/tex]:
Using polynomial long division or factoring, we find:
[tex]\[
3x^3 + 18x^2 + 27x = 3x(x+3)^2
\][/tex]
Therefore:
[tex]\[
3x(x+3)^2 \div (x+3) = 3x(x+3)
\][/tex]
The result, [tex]\(3x(x+3)\)[/tex], is a polynomial. Thus, [tex]\(x+3\)[/tex] is a factor of [tex]\(3x^3 + 18x^2 + 27x\)[/tex].
4. Testing [tex]\(x-3\)[/tex]:
Dividing [tex]\(3x^3 + 18x^2 + 27x\)[/tex] by [tex]\(x-3\)[/tex]:
Using polynomial long division or substitution method to check if [tex]\(x-3\)[/tex] is a factor:
[tex]\[
3(3)^3 + 18(3)^2 + 27(3) = 3(27) + 18(9) + 27(3) = 81 + 162 + 81 = 324 \neq 0
\][/tex]
Since plugging [tex]\(x = 3\)[/tex] into the polynomial does not yield zero, [tex]\(x-3\)[/tex] is not a factor of [tex]\(3x^3 + 18x^2 + 27x\)[/tex].
In conclusion, the factors of the polynomial [tex]\(3x^3 + 18x^2 + 27x\)[/tex] from the given choices are:
[tex]\[ 9x \quad \text{and} \quad x+3 \][/tex]
