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