Answer: 129
Step-by-step explanation:
If a polynomial has zeros at x = a, x = b, and x = c, then the polynomial can be expressed as:
\($p(x) = (x - a)(x - b)(x - c)$\)
Given the zeros 6, 5, and 9, the polynomial becomes:
\($p(x) = (x - 6)(x - 5)(x - 9)$\)
We need to expand this product to find the polynomial in standard form
First expand \((x-5)(x-6)\):
\((x-6)(x-5)=x^2-5 x-6 x+30=x^2-11 x+30\)
Now, multiply this result by \((x-9):\)
\((\mathrm{x}^2-11 \mathrm{x}+30\right)(\mathrm{x}-9)=\mathrm{x}^3-9 \mathrm{x}^2-11 \mathrm{x}^2+99 \mathrm{x}+30 \mathrm{x}-270\)
Combine like terms:
\(x^3-20 x^2+129 x-270\)
Therefore, the polynomial in standard form is:
\(p(x)=x^3-20 x^2+129 x-270\)
The coefficient of the x term in this polynomial is 129
