X^4+x^3-6x^2-14x-12=0 Make a list of possible rational roots. Test the possible roots until you find one that produces a remainder of 0 Write the resulting cubic function. Use synthetic division to find a second root that will reduce the cubic expression to a quadratic expression

The Rational Roots Test states that for a polynomial with integer coefficients, the factors of the constant / the factors of the leading coefficient are the possible rational roots.

Here, the constant (the value without an x attached to it) is -12 and the leading coefficient (the value that the x to the highest degree is multiplied by) is 1 as x⁴ is multiplied by 1. The factors of -12 are

±(1, 2, 3, 4, 6, 12), so the possible rational roots are ±(1, 2, 3, 4, 6, 12)/1 (as 1 is the only factor of 1).

Trying out a few roots until we get one that works using synthetic division, we can try

x+1 (the root is x=-1)

-1 | 1 1 -6 -14 -12

| -1 0 6 8

__________________________

1 0 -6 -8 -4

the remainder is -4, so this does not work

x+2 (the root is x=-2)

-2 | 1 1 -6 -14 -12

| -2 2 8 12

__________________________

1 -1 -4 -6 0

Therefore, x=-2 is a root and x+2 is a factor of the polynomial. The quotient of the polynomial and x+2 is

-6 + (-4)x + (-1)* x² + 1 * x³ = x³-x²-4x-6

Using the rational roots theorem, the possible roots of x³-x²-4x-6 are

±(1,2,3,6)

Starting with

x-1 (root is x=1), we have

1 | 1 -1 -4 -6

| 1 0 -4

_____________________

1 0 -4 -10

there is a remainder, so this is not a root

next, x-2 (root is x=2)

2 | 1 -1 -4 -6

| 2 2 -4

_____________________

1 1 -2 -10

there is a remainder, so this is not a root

next, x-3 (root is x=3)

3| 1 -1 -4 -6

| 3 6 6

_____________________

1 2 2 0

x-3 is a factor and 3 is a root. the quotient of (x³-x²-4x-6)/(x-3) is x²+2x+2