To determine which of the provided options is not a factor of the polynomial \(3x^3 - 10x^2 - 143x - 90\), we can use polynomial long division to divide the polynomial by each of the possible factors.
Let's test each of the options one at a time:
- Divide by \(x - 9\)
Let's perform the division of \(3x^3 - 10x^2 - 143x - 90\) by \(x - 9\).
\[ \begin{array}{r|rrrr} & 3x^2 & 17 & 2 \ x - 9 & 3x^3 & -10x^2 & -143x & -90 \ & -(3x^3 - 27x^2) \ \hline & 0 & 17x^2 & -143x \ & -(17x^2 - 153x) \ \hline & 0 & 10x & -90 \ & -(10x - 90) \ \hline & 0 & 0 \ \end{array} \]
The remainder is \(0\), so \(x - 9\) is a factor.
- Divide by \(3x + 2\)
Now let's divide \(3x^3 - 10x^2 - 143x - 90\) by \(3x + 2\).
\[ \begin{array}{r|rrrr} & x^2 & -4 & -13 \ 3x + 2 & 3x^3 & -10x^2 & -143x & -90 \ & -(3x^3 + 2x^2) \ \hline & 0 & -12x^2 & -143x \ & -(12x^2 + 8x) \ \hline & 0 & -135x & -90 \ & -(-135x - 90) \ \hline & 0 & 0 \ \end{array} \]
The remainder is \(0\), so \(3x + 2\) is a factor.
- Divide by \(x + 5\)
Next, let's divide \(3x^3 - 10x^2 - 143x - 90\) by \(x + 5\).
\[ \begin{array}{r|rrrr} & 3x^2 & -25 & 7 \ x + 5 & 3x^3 & -10x^2 & -143x & -90 \ & -(3x^3 + 15x^2) \ \hline & 0 & -25x^2 & -143x \ & -(-25x^2 - 125x) \ \hline & 0 & -18x & -90 \ & -(-18x - 90) \ \hline & 0 & 0 \ \end{array} \]
The remainder is \(0\), so \(x + 5\) is a factor.
- Divide by \(x - 6\)
Finally, let's divide \(3x^3 - 10x^2 - 143x - 90\) by \(x - 6\).
\[ \begin{array}{r|rrrr} & 3x^2 & 8 & -7 \ x - 6 & 3x^3 & -10x^2 & -143x & -90 \ & -(3x^3 - 18x^2) \ \hline & 0 & 8x^2 & -143x \ & -(8x^2 - 48x) \ \hline & 0 & -95x & -90 \ & -(-95x + 570) \ \hline & 0 & 0 \ \end{array} \]
The remainder is \(570\), which is not \(0\). Therefore, \(x - 6\) is not a factor of the polynomial.
Conclusion
The option that is not a factor of \(3x^3 - 10x^2 - 143x - 90\) is:
D. \(x - 6\).