William P. answered • 01/14/20

University Math Instructor and Experienced Calculus Tutor

Hello Michelle,

To find the rational zeros of the polynomial function f(x) = 3x^{4} + 16x^{3} - 59x^{2} - 400x - 400 (equivalently, to solve the equation 3x^{4} + 16x^{3} - 59x^{2} - 400x - 400 = 0), we begin by using the Rational Zeros Theorem:

**Rational Zeros Theorem**

Let f(x) be a polynomial function with integer coefficients. If the rational number p/q (where p and q have no common factors other than ±1) is a zero of f(x), then p is a factor of the constant term of f(x) and q is a factor of the lead coefficient of f(x).

As applied to your problem, this says that the **possible** rational zeros of

f(x) = 3x^{4} + 16x^{3} - 59x^{2} - 400x - 400

will have the form p/q, where p is a factor of -400 and q is a factor of 3. First, list all the integer factors of the constant term -400.

p: ±1,±2,±4,±5,±8,±10,±16,±20,±25,±40,±50,±80,±100,±200,±400

The factors of the lead coefficient 3 are

q:±1,±3

The **possible** rational zeros of the polynomial are all numbers of the form p/q.

p/q: ±1,±2,±4,±5,±8,±10,±16,±20,±25,±40,±50,±80,±100,±200,±400,

±1/3, ±2/3, ±4/3, ±5/3, ±8/3, ±10/3, ±16/3, ±20/3, ±25/3, ±40/3, ±50/3, ±80/3, ±100/3,±200/3, ±400/3

To determine which (if any) of these possible rational zeros are actually zeros, we can use **synthetic division.** Recall that if you divide a polynomial f(x) by x - r and the remainder is 0, then f(r) = 0 (that is, r is a zero of f(x).) Let us check if 1 is a zero by synthetic division.

1 _| 3 16 -59 -400 -400 (Coefficients Row)

__ 3 3 -2 -402__ ("Work Row")

3 19 -2 -402 **-802** **<----Remainder** ("Results Row")

The remainder is -802 (not zero), so 1 is not a zero of the polynomial. There are a total of 60 possible rational zeros in the list, and it would take a long time to check them all. I will skip ahead and try 5 next.

5 _| 3 16 -59 -400 -400

__ 15 155 480 400__

3 31 96 80 0

Since the remainder is zero, **5 is a zero of f(x)**. From the results of this division, we know that x - 5 is a factor of f(x), and the quotient (result of the division) is 3x^{3} + 31x^{2} + 96x + 80. That is, f(x) may be written as

f(x) = (x - 5)(3x^{3} + 31x^{2} + 96x + 80).

The remaining zeros of f(x) must be zeros of 3x^{3} + 31x^{2} + 96x + 80, so we now carry out synthetic division on this polynomial. Synthetic division by -4 gives a remainder of 0, so **-4 is also a zero**. The quotient obtained is 3x^{2} + 19x + 20, and the remaining zeros of f(x) must be zeros of this polynomial. That is, they must be solutions of the equation 3x^{2} + 19x + 20 = 0. You can solve this using the quadratic formula. The solutions are complex numbers, so there are no more rational roots. Therefore, we conclude (finally) that** the rational roots of f(x) are 5 and -4**. Hope that helps! Let me know if you have any questions.

William

William P.

01/14/20

William P.

01/14/20

Doug C.

Nice presentation. Just commenting because I was working on this when I noticed your answer. That last quadratic does in fact factor. the two additional rational roots are -5 and -4/3.01/14/20