Question 64288
The given expression is: 


f(x) = {{{(x+1)^3 (2x + 3)(2 - x)^2}}} 


This can be written as:



f(x) = (x+1)(x+1)(x+1)(2x+3)(2-x)(2-x) --------------(EQN 2) 


So, on multiplying the given expression it can be writtne as: 


{{{(x^3 + 3x^2 + 3x + 1)(2x + 3)(x^2 - 4x + 4)}}} 


This can be simplified and reduced and written as: 


{{{(2x^6 + x^5 - 13x^4 - 13x^3 + 19x^2 + 32x + 12)}}} 


Hence, the highest power of x gives us the degree of the polynomial. 



The degree of f(x) = 6 



Now, the zeros of this one is obtained by: 



Equating (2) to zero.


Thus, we find that the zeros of the polynomial are: 


x = -1, -1, -1, -2/3, 2, 2



Hence, the solution.


Regards...