# Math Polynomials Help?

x^4 - 4x^3 - x^2 + 16x -12 using the factor theorem

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• What's the question?  Factor?

If so we would need to find two rational roots to that to turn them into factors.

The list of possible rational roots can be made by checking all factors of the constant term and dividing them by all factors of the high-degree coefficient.  Since the latter is 1, that makes things easier.  This list based on the constant term is:

±1, ±2, ±3, ±4, ±6, ±12

There are 12 possible outcomes to check.  Let's test the 1's first:

x⁴ - 4x³ - x² + 16x - 12

(-1)⁴ - 4(-1)³ - (-1)² + 16(-1) - 12 and 1⁴ - 4(1)³ - 1² + 16(1) - 12

1 - 4(-1) - 1 + 16(-1) - 12 and 1 - 4(1) - 1 + 16(1) - 12

1 + 4 - 1 - 16 - 12 and 1 - 4 - 1 + 16 - 12

-24 and 0

We have one root: x = 1, which makes (x - 1) a factor.  We need one more, so checking the 2's:

x⁴ - 4x³ - x² + 16x - 12

(-2)⁴ - 4(-2)³ - (-2)² + 16(-2) - 12 and 2⁴ - 4(2)³ - 2² + 16(2) - 12

16 - 4(-8) - 4 + 16(-2) - 12 and 16 - 4(8) - 4 + 16(2) - 12

16 + 32 - 4 - 32 - 12 and 16 - 32 - 4 + 32 - 12

0 and 0

Both of these are roots: x = -2 and x = 2, so the factors are (x + 2) and (x - 2)

We have three factors now.  The product of the factors will also be a factor, so let's expand that out:

(x - 1)(x + 2)(x - 2)

(x - 1)(x² - 4)

x³ - x² - 4x + 4

We can now divide the original polynomial by this cubic to get the final factor.  If we only found two factors the end result would be a quadratic, which can then be solved in many ways and do not have to be rational at this point:

. . . . . . . . . . . . _x_-_3______________

x³ - x² - 4x + 4 ) x⁴ - 4x³ - x² + 16x - 12

. . . . . . . . . . . . . x⁴ - x³ - 4x² + 4x

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

. . . . . . . . . . . . . . . -3x³ + 3x² + 12x - 12

. . . . . . . . . . . . . . . -3x³ + 3x² + 12x - 12

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

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

Now we have the final factor:

x⁴ - 4x³ - x² + 16x - 12

(x - 3)(x - 2)(x - 1)(x + 2)