H(x) = x^(2) + 43x +120 find the zero of the functions

Question

`h(x) = x^(2) + 43x +120`
find the zero of the functions

Answer 1

To solve this set it equal to 0

`0 = {x}^(2) + 43x + 120`
Now factor

`0 = (x + 40)(x + 3)`
Set each side to 0 and solve

`x + 40 = 0 \\ x = - 40 \\ x + 3 = 0 \\ x = - 3`
So the zeros (where it crosses the x axis) are

`x = - 40 \\ x = - 3`