Question

Solve for x

x^4 + 2x^3 + 2x^2 + 3x + 1 = 0

Answer 1

As per the Descartes’ rule of signs, the number of positive real zeroes in a polynomial function f(x) is the same as the number of changes in the sign of the coefficients. or less than this number by an even number. Similarly, the number of negative real zeroes of the polynomial function f(x) is the same as the number of changes in sign of the coefficients of the terms of f(-x) or less than this number by an even number.

Here, f(x) = x⁴ + 2x³ + 2x² + 3x+ 1 . We may observe that there is no change of sign of the coefficients of f(x). Hence, f(x) has no real positive zeros. Further, f(-x) = x⁴ - 2x³ + 2x² - 3x+ 1. We may observe that there are 4 changes of sign here so that there may be 4 or 2 or no real negative zeros.

Since it is difficult to factorize a 4^th degree polynomial, hence the best way to determine its zeros is to make a graph. The points where the graph of f(x) crosses the X-Axis are the zeros of f(x) .

A graph of f(x) prepared by using the Desmos graphing calculator is attached. It may be observed that f(x) has 2 real zeros at -1.664 and -0.408. ( The coordinates may be seen in the original Desmos graph). The other 2 zeros of f(x) are a conjugate pair of complex numbers as the graph of f(x) crosses the X-Axis at only 2 points.

If the exact values, say -a and -b of the real zeros were available, we would have divided f(x) by (x+a)(x+b) to arrive at a quadratic polynomial. Then, on using the quadratic formula, we would have determined the 2 complex zeros of f(x). Here, since the exact values of the real zeros of f(x) are difficult to determine the 2 complex zeros of f(x).