Roots of a polynomial

Question

How would I find the zeros of the polynomial g(x)=x^4-2x^3-3x^2+5x+2.

I've simplified the expression to (x-2)(x^3-3x-1), am I missing something?

Answer 1

Not sure what the expectation is but perhaps you are supposed to graph the residual polynomial factor

(x³- 3x - 1) and determine where that function crosses the x-axis). There are 3 real solutions to the equation

x³-3x -1 = 0.

They are:

x₁≈-1.532

x₂≈-0.347

x₃≈1.879

Since you are not in Calculus (yet), you would not be aware of something called Newton's method for approximating roots of a polynomial function with real solutions.

There are also some very complicated formulas for solving cubic equations, but to use them requires an advanced understanding of operations with complex numbers.

desmos.com/calculator/slholublbd

The above is a Desmos graph that shows the above roots and also shows the use of Newton's method, something you are a few years away from learning about.

Answer 2

http://www.math.utah.edu/~wortman/1060text-tcf.pdf shows how to solve this special cubic.

Following the steps laid out here:

a = -3; b = -1

D = -b/2 + i sqrt(-(a/3)^3-(b/2)^2) = -b/2 + i sqrt(-(a/3)^3 - (b/2)^2) = 1/2 + i sqrt(3)/4

D is familiar to us. It is cos 60° + i sin 60º

Thus, D^(1/3) = cos 20° + i sin 20º

Thus, R = cos 20° and I = sin 20°

Our solutions are 2R, -R + √3 I, -R - √3 I

Thus, the solutions are 2 cos 20°, -1 cos 20° + √3 sin 20°, -1 cos 20° - √3 sin 20°

(x - 2) (x - 2 cos 20°) (x + cos 20° - √3 sin 20°) (x + cos 20° + √3 sin 20°)

Answer 3

Nothing.

2 is a zero.

Do you know the Rational Root Theorem?

The possible roots are ±1

f(1) = 1³ - 3 - 1, is that zero?

f(-1) = -1 + 3 - 1, is that zero?