Answer:
The zeros of the function f(x) = 9x^3 - 45x^2 + 36x is 0, 1, 4
Solution:
Given that
![f(x)=9 x^(3)-45 x^(2)-36 x](https://img.qammunity.org/2020/formulas/mathematics/middle-school/wdfiuf3yv1hgt1gnn52pftn2xxcx6n6z7o.png)
For finding the zeros of the function, we equate the entire function to zero i.e.,
![0=9 x^(3)-45 x^(2)+36 x](https://img.qammunity.org/2020/formulas/mathematics/middle-school/zjrg1gfupp00zxj3inuiqyjo8wc5kpnjbo.png)
Dividing throughout by 9, we get
![0=x^(3)-5 x^(2)+4 x](https://img.qammunity.org/2020/formulas/mathematics/middle-school/wzci6v2733znz6y825p3gyflbu06061sst.png)
Taking x as common throughout the equation, we get
![0=x\left(x^(2)-5 x+4\right)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/xourkltncm5c7r15wbf1yyxhectrjbfm5u.png)
Thus, by factorization of the above equation, we get 0 = x(x - 1)(x - 4)
Now ,equating the factors we got to 0, we get
x = 0, x - 1 = 0, x - 4 = 0
x = 0, x = 1, x = 4
Thus, the zeros of the above given function are 0, 1, 4