Question

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?

Answer 1

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]}