Sagot :
The way i do it is factor x^2-12x+32 into (x-8)(x-4) then set each of those to zero which would look like 0=x-8 and 0=x-4. Then solve each equation and your answers would be x=8 and x=4
Answer:
The zeros are x=4, 8
Step-by-step explanation:
We have been given the function y=x^2-12x+32.
Let us write this in intercept form by factoring the given function.
We can factor it by AC method.
We can write the middle term -12x = -8x-4x
[tex]y=x^2-12x+32\\y=x^2-8x-4x+32\\\text{Now we take GCF}\\\\y=x(x-8)-4(x-8)\\\\y=(x-8)(x-4)[/tex]
Now, in order to find the zeros, we have
[tex](x-8)(x-4)=0\\\\x=4,8[/tex]
Therefore, the zeros are x=4, 8