How do we find the number of distinct real roots of the equation x^4-4x^3+12x^2+x-1=0 without using graph?

1 Answer
Aug 11, 2018

This quartic has exactly two real roots.

Explanation:

Given:

x^4-4x^3+12x^2+x-1 = 0

Let:

f(x) = x^4-4x^3+12x^2+x-1

Note that the pattern of signs of the coefficients is + - + + -. With 3 changes of sign, Descartes' Rule of Signs tells us that f(x) has 3 or 1 positive real zeros.

The pattern of signs of the coefficients of f(-x) is + + + - -. With exactly one change of signs, Descartes' Rule of Signs tells us that f(x) has exactly one negative real zero.

Also note that:

x^4-4x^3+12x^2+x-1

= (x-1)^4 + 6(x-1)^2 + 17(x-1) + 9

Note that the coefficients of all the expressions in (x-1) on the right hand side are positive. So there is no x >= 1 for which this is zero.

Hence we can deduce that any real zeros of x^4-4x^3+12x^2+x-1 are in (0, 1).

f'(x) = 4x^3-12x^2+24x+1

color(white)(f'(x)) = 4(x^3-3x^2+3x-1)+12(x-1)+17

color(white)(f'(x)) = 4(x-1)^3+12(x-1)+17

which is positive for all x in (0, 1)

Hence f(x) can only have one zero in (0, 1), making a total of 2 real zeros including the negative one.

Now let's see the graph...

graph{x^4-4x^3+12x^2+x-1 [-1.2, 1.2, -1.25, 1.25]}