h(x)=x^4 - 9x^3 + 18x^2WHAT ARE THE ZEROS OF THE FUNCTION

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.