h(x) =x^4-9x^3+18x^2

To find the zeros of f(x) we first factorise the expression on the left of the equation and then equate each factor to zero and solve for x.

x^4-9x^3 +18 x^2 = 0

Here x^2 is a factor of each term . So we factor out x^2.

x^2(x^2-9x+18) = 0 ......(1).

Now we consider x^2-9x+18 for factorisation.

x^2 -9x +18 = x^2-6x-3x +18 = 0

x(x-6) -3(x-6) = 0

(x-6)(x-3) = 0.

Therefore x^4-9x^3+18x^2 = x^2(x-6)(x-3) = 0

Equate each factor to zer and solve for x.

x^2 = 0 gives x = 0

x-6 = 0 gives x = 6

x-3 = 0 gives x = 3.

So x= 0, x = 6 x = 3 are the solutions for x.

The zeroes of a function are values of x for which the values of the function becomes equal to zero. Here we have the function h(x)=x^4 - 9x^3 + 18x^2 . To find the zeroes, we equate it to zero.

x^4 - 9x^3 + 18x^2 =0

=> x^2 (x^2 - 9x + 18) =0

=> x^2 ( x^2 - 6x - 3x +18 ) =0

=> x^2 [ x(x-6) - 3(x-6)] =0

=> x^2 [(x-6)(x-3)] =0

=> x^2 (x-6) (x-3) =0

So either x =0 or x - 6 =0 or x - 3 =0

Therefore x can be equal to 0 , 6 or 3.

**The zeroes of the function are 0, 3 and 6.**