Question

Find all real zeros h(x)=5(x-5)(x+5)^2(x^2-1)

Answer 1

`x=-5,-1,1,5`

Explanation:

Use the Zero Product Property and set each factor equal to 0

`h(x)=5(x-5)(x+5)^2(x^2-1)`

`0=x-5\\x=5`

`0=(x+5)^2\\0=x+5\\x=-5`<-- Multiplicity of 2

`0=x^2-1\\0=(x-1)(x+1)\\x=-1,1`

Therefore, all real zeroes of the function are
`x=-5,-1,1,5`