Verified answer

The first thing we need is to find one rational zero (if they exist), then from there we can find the other two.

A list of possible rational roots can be created by listing the factors of the constant term and dividing them by the factors of the high-degree coefficient.

Since the latter is 1, that makes things easy and leaves the constant. Since it's 4, that makes the list of possible rational roots:

±1, ±2, ±4.

So, let's start checking if any are zeroes. With that 9 in there, I doubt the 1's are, so let's start with the 2's:

P(x) = x³ - 2x² - 9x + 4

P(-2) = (-2)³ - 2(-2)² - 9(-2) + 4 and P(2) = (2)³ - 2(2)² - 9(2) + 4

P(-2) = -8 - 2(4) + 18 + 4 and P(2) = 8 - 2(4) - 8 + 4

P(-2) = -8 - 8 + 22 and P(2) = 8 - 8 - 4

And from here, I can see neither are. Let's test the 4's:

P(x) = x³ - 2x² - 9x + 4

P(-4) = (-4)³ - 2(-4)² - 9(-4) + 4 and P(4) = (4)³ - 2(4)² - 9(4) + 4

P(-4) = -64 - 2(16) + 36 + 4 and P(4) = 64 - 2(16) - 36 + 4

P(-4) = -64 - 32 + 40 and P(4) = 64 - 32 - 32

P(-4) = -56 and P(4) = 0

So 4 is a root of P(x), so (x - 4) is a factor. Knowing this we can divide the cubic by the factor to get a quadratic, which then we can solve no matter what types or roots it has:

. . . ._x²_+_2x_-_1_____

x - 4 ) x³ - 2x² - 9x + 4

. . . . .x³ - 4x²

. . . ----------------

. . . . . . . 2x² - 9x + 4

. . . . . . . 2x² - 8x

. . . . . . ---------------

. . . . . . . . . . . -x + 4

. . . . . . . . . . . -x + 4

. . . . . . . . . . -------------

. . . . . . . . . . . . . . 0

So now we can factor this to:

x³ - 2x² - 9x + 4 = 0

(x - 4)(x² + 2x - 1) = 0

We know x - 4 = 0 already, so we just have to solve for x² + 2x - 1 = 0:

x² + 2x - 1 = 0

completing the square by adding 2 to both sides:

x² + 2x + 1 = 2

(x + 1)² = 2

x + 1 = ± √2

x = -1 ± √2

So your three real roots are:

x = 4 and (-1 ± √2)