Question

How do you find all the zeros of `f(x)=x^3 + 4x^2 - 5`?

Answer 1

`x_1=1` and `x_2=(-5+sqrt5)/2` and `x_3=(-5-sqrt5)/2`

Explanation:

From the given:
`y=x^3+4x^2-5`

Perform synthetic division

`x^3" " " "x^2" " " "x^1" " " " "x^0`

`1" " " " " 4" " " " "0" " " -5` trial divisor `=1`
`underline(" " " " " " "1" " " " " "5" " " " " "5)`
`1" " " " " 5" " " " " "5" " " " " " 0` `larr`remainder

`x_1=1` is a zero

depressed equation:

`x^2+5x+5=0`

by quadratic formula

`x=(-b+-sqrt(b^2-4ac))/(2a)`

Let `a=1` and `b=5` and `c=5`

`x=(-b+-sqrt(b^2-4ac))/(2a)`

`x=(-5+-sqrt(5^2-4(1)(5)))/(2(1))`

`x=(-5+-sqrt5)/2`

the other zeros are:
`x_2=(-5+sqrt5)/2`

`x_3=(-5-sqrt5)/2`

God bless....I hope the explanation is useful.