How do you take a factor out of polynomial?

Question

The problem I have in particular in  f(x) = 3x^3 - 5x^2 - 58x + 40

given that f(3) = -198

and f(5) = 0

find all the solutions of f(x) = 0

Answer 1

Factor out x - 5, since 5 is a zero of this polynomial.

                       3x²   +   10x   -    8

x - 5 )  3x³   -   5x²    -   58x   +   40

          3x³   -   15x²

                      10x²   -   58x

                      10x²   -   50x

                                    -8x   +   40

                                    -8x   +   40

                                                  0

The other answers are the zeroes of 3x² + 10x - 8.

The two numbers with a sum of b = 10 and a product of ac = 3*(-8) = -24 are 12 and -2 so you can factor as follows.

3x² + 10x - 8 = 3x² + 12x - 2x - 8 = 3x(x+4) - 2(x+4) = (3x - 2)(x + 4)

x + 4 = 0 gives x = -4.

3x - 2 = 0 gives x = 2/3.

So the solution set is {-4, 2/3, 5}.

Answer 2

If you have a function f(x), where f(a) = 0, then (x - a) is a factor of f(x).

Because f(5) = 0, (x - 5) is a factor of f(x).

Using synthetic division, we can divide f(x) by (x - 5) and find the remaining factor.

5| 3  -5  -58  40
  |     15  50  -40
     3  10  -8     0

The other factor is 3x² + 10x - 8

This can be factored as (3x - 2)(x + 4)

By setting each factor equal to zero, we can find the each of the solutions for f(x) = 0.

3x - 2 = 0                      x + 4 = 0

3x = 2                          x = -4

x = 2/3

Since we already know that f(5) = 0, 5 is also a solution.

x = {-4, 2/3, 5}