Answer:
is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.
Explanation:
Given the function
![f\left(x\right)=x^3-6x^2+3x+10](https://img.qammunity.org/2021/formulas/mathematics/high-school/mqec8dgqpkap361t846h2g6s5jsxoc3qi8.png)
As the highest power of the x-variable is 3 with the leading coefficients of 1.
- So, it is clear that the polynomial function of the least degree has the real coefficients and the leading coefficients of 1.
solving to get the zeros
![f\left(x\right)=x^3-6x^2+3x+10](https://img.qammunity.org/2021/formulas/mathematics/high-school/mqec8dgqpkap361t846h2g6s5jsxoc3qi8.png)
∵
![f(x)=0](https://img.qammunity.org/2021/formulas/mathematics/high-school/wvi3mxzpky8dxx75wjj4eulze2k1h1l2kh.png)
as
![Factor\:x^3-6x^2+3x+10\::\:\left(x+1\right)\left(x-2\right)\left(x-5\right)=0](https://img.qammunity.org/2021/formulas/mathematics/high-school/fhbp1gu240h7a17p5egm567a92cbk56k4d.png)
so
Using the zero factor principle
if
![ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)](https://img.qammunity.org/2021/formulas/mathematics/college/pgix1ewdj003su3mtaw6bkaotoeft5gm62.png)
![x+1=0\quad \mathrm{or}\quad \:x-2=0\quad \mathrm{or}\quad \:x-5=0](https://img.qammunity.org/2021/formulas/mathematics/high-school/aady22si92vdu6nunu28dl1nlpocc17wn4.png)
![x=-1,\:x=2,\:x=5](https://img.qammunity.org/2021/formulas/mathematics/high-school/7nifyteqsmvlbm4we4cw7sokd3kyouzp5k.png)
Therefore, the zeros of the function are:
![x=-1,\:x=2,\:x=5](https://img.qammunity.org/2021/formulas/mathematics/high-school/7nifyteqsmvlbm4we4cw7sokd3kyouzp5k.png)
is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.
Therefore, the last option is true.