With the following polynomials their zeroes are given. Find their all other zeroes:

f(x) = x³ + 13x² + 32x + 20; –2

Given that -2 is the zero of the polynomial.

So x = -2 which is x + 2 = 0

The other zeroes are as follows:

x² + 11x + 10 = 0

Solving the above quadratic equation.

Sum = 11

Product = 10

So the numbers which satisfy the above condition are 10 and 1

x² + 10x + x + 10 = 0

x(x + 1)+ 10(x + 1) = 0

(x + 10) (x + 1) = 0

Solving the first part,

x + 10 = 0

x = -10

Solving the second part,

x + 1 = 0

x = -1

Therefore the other zeroes of the polynomials are -1 and -10.