Solve the Polynomial equation.

Question

Use the Rational Zero Theorem, Descartes's Rule of​ Signs, and possibly the graph of the polynomial function shown by a graphing utility as an aid in obtaining the first root.

x^4-4x^3-57x^2-62x-10=0

Answer 1

x⁴ - 4x³ - 57x² - 62x - 10 = 0

Because this is a 4th degree polynomial, by the Fundamental Theorem of Algebra, we know it has 4 roots.

By the Rational Roots Theorem, we know that if there are any rational roots, they will be ±10, ±5, ±2, or ±1.

By Descarte's Rule of Signs, the polynomial will have 1 positive real root (one sign change in the equation) and will have either 3 or 1 negative real roots (meaning it will have either 1 or 3 complex roots).

So let's start by trying to find the positive real root. Since the coefficients are a bit high, I'm going to start with 10 as my first guess. Doing synthetic division with "10" to get:

10 | 1 -4 -57 -62 -10

10 60 30 -320

---------------------------------

1 6 3 -32 -330

Notice that we did NOT get zero so 10 is not a root and we did also NOT get the tell-tale sign that 10 is an upper bound (meaning the root is higher than 10). This tells me that the one positive real root, which must be greater than 10, will not be rational. So, I'll skip it for now.

Trying negative rational roots, let's try -1:

-1 | 1 -4 -57 -62 -10

-1 5 52 10

---------------------------------

1 -5 -52 -10 0

So, x = -1 is a root.

Let's try x = -2 but we'll try it using the reduced polynomial that we got from the synthetic division of x = -1 which is x³ - 5x² - 52x - 10:

-2 | 1 -5 -52 -10

-2 14 76

-----------------------

1 -7 -38 66

So, x = -2 is not a root. It is also not a lower bound, so let's try -5:

-5 | 1 -5 -52 -10

-5 50 10

---------------------------------

1 -10 -2 0

So x = -5 is a root. and the resulting quadratic is x² - 10x - 2 = 0. Since this is not factorable, use the quadratic formula:

x = [10 ± √(100 - 4(-2))]/2 = (10 ± √108)/2 = (10 ± 6√3)/2 = 5 ± 3√3

Answer 2

By Rational Zeros Theorem list of possible rational zeros: ±1, ±2, ±5, ±10.

By trying or uzing TI-84 graph we get: - 1 and - 5 are zeros.

Tom get other zeros and check if we get right we use 2-times synthetic divisio

-5 | 1 -4 -57 -62 -10

-5 45 60 10

-1 | 1 -9 -12 -2 | 0

-1 10 2

1 -10 -2 |0

We get quotient and set it to 0: x² - 10x - 2 = 0; x = 5 ± √27 = 5 ± 3√2.

So solution set: {-5, -1, 5 - 3√3, 5 + 3√3}