A polynomial equation of degree x^4 + 2x^3 − 3x^2 − 28x − 24 = 0

Question

Use the rational root theorem and the factor theorem to help solve the equation. Be sure that the number of solutions for the equation agrees with the property below, taking into account multiplicity of solutions. (Objectives 1, 2)

A polynomial equation of degree n has n solutions, and any solution of multiplicity p is counted p times.

x⁴ + 2x³ − 3x² − 28x − 24 = 0

 

Answer 1

x^4 + 2x^3 - 3x^2 - 28x - 24 = 0

The factors of 24 are: 1, -1, 2, -2, 3, -3, 4, -4, 6, -6, 8, -8, 12, -12, 24, -24
The factors of 1 are: 1, -1

Note that 24 is the constant, and 1 is the coefficient of x^4

By the rational root theorem, if a root is rational, it will have to be of the form:
(factor of 24) / (factor of 1).

Thus, the possible rational roots are: 1, -1, 2, -2, 3, -3, 4, -4, 6, -6, 8, -8, 12, -12, 24, -24

You can try substituting the possible roots in till you get a correct root.
The correct roots will be x = -1 and x = 3 The factors will thus be (x + 1) and (x - 3).

Using long division on x^4 + 2x^3 - 3x^2 - 28x - 24 with (x + 1), and then with (x - 3) a second time, you will have x^2 + 4x + 8, which has no real roots.

Hence, x = -1 and x = 3 are the only real roots.