Question

How do you find all the zeros of `f(x)=x^3 -8x^2 -23x +30`?

Answer 1

They are `x=1, x=10, x=-3`.

Explanation:

To solve a cubic equation it exists the solving formula but it is very long and I do not know it.
Then I use a bit of luck to solve this equation. For example I see that `x=1` is a solution because

`1^3-8*1^2-23*2+30=1-8-23+30=0`.

Then I know that the equation has to be in the form

`(ax^2+bx+c)(x-1)`

To find `a, b, c` I just do the multiplication and compare the coefficients.

`(ax^2+bx+c)(x-1)`

`=ax^3+bx^2+cx-ax^2-bx-c`

`=ax^3+(b-a)x^2+(c-b)x-c`

Comparing this with our initial equation we see that

`a=1`
`b-a=-8, b-1=-8, b=-7`
`c-b=-23, c+7=-23, c=-30`
`-c=30, c=-30`

Our function is then

`(x^2-7x-30)(x-1)`.

Now we have to solve a second order equation `(x^2-7x-30)` using the solution

`x=(-b\pmsqrt(b^2-4ac))/(2a)`

`=(-(-7)\pmsqrt((-7)^2-4*1*(-30)))/2`

`=(7\pmsqrt(49+120))/2`

`=(7\pmsqrt(169))/2`

`=(7\pm13)/2`

that has the two solutions

`x=(7+13)/2=10` and `x=(7-13)/2=-3`.

Then the function

`f(x)=x^3-8x^2-23x+30`

can be rewritten as

`f(x)=(x-1)(x-10)(x+3)` with the three zeros for `x` equal to `1, 10, -3`.