Question

If f(x)=x^3-8x^2-23x+30 and x+3 is a factor if f(x), then find all of the zeros of f(x) algebraically!

Answer 1

Finding the Zeros of a Cubic Function given one Factor

Answer:

`f(x) = 0 \text{ when } x \in \{ 1, -3, 10 \}`

Explanation:

Given:

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

Note that if
`x +3` is a factor of
`f(x)`, then
`(x +3)a = f(x)` where
`a` is also a factor of
`f(x)`. This is the same logic as
`\blue 3` is a factor of
`12` so
`\blue 3 * a = 12` and
`a` would be
`4` which we got it by dividing
`12` by
`\blue 3`. In this case, we can divide
`f(x)` by
`x +3` to get the other factor of
`f(x)` (The solution for that is found on the attached image).

`a = x^2 -11x +10`

We also have to note that multiplying any number with
`0` gives
`0`. If the value of
`x` makes
`x +3` equal
`0`,
`(x +3)a` will be
`0` no matter what the value of
`a` will be. We can see that
`x = -3` makes
`x +3 = 0`.
`x = -3` is one of our zeros.

Now we have to solve at what value of
`x` make
`a = 0`.

Solving for
`x`:

`a = 0 \\ x^2 -11x +10 = 0 \\ x^2 -11x = -10 \\ x^2 -11x +(121)/(4) = -10 +(121)/(4) \\ (x -(11)/(2))^2 = -(40)/(4) +(121)/(4) \\ (x -(11)/(2))^2 = (81)/(4) \\ x -(11)/(2) = \pm \sqrt{(81)/(4)} \\ x -(11)/(2) = \pm (9)/(2)`

Solving from the Positive Root:

`x -(11)/(2) = (9)/(2) \\ x = (9)/(2) +(11)/(2) \\ x = (20)/(2) \\ x = 10`

Solving from the Negative Root:

`x -(11)/(2) = - (9)/(2) \\ x = -(9)/(2) +(11)/(2) \\ x = (2)/(2) \\ x = 1`