Question

What are the zeros of f(x) = x2 - 12x + 36?

Answer 1

Answer:

c. x = 6 only

Step-by-step explanation:

In order to calculate the zeros of f(x), we need to set it equal to zero and find the corresponding values of x.

$x^{2}-12x+36=0$

Using the midterm breaking, we can split -12x into two such terms whose sum will be -12x and product will be 36x². These two terms are -6x and -6x

So, the above expression can be written as:

$x^2-6x-6x+36=0\\\\ x(x-6)-6(x-6)=0\\\\ (x-6)(x-6)=0\\\\ (x-6)^{2}=0\\\\ x-6=0\\\\ x=6$

This means, the zero of f(x) occurs at x = 6 only.