How do i find the "EXACT" no of real roots of a equation?No HIT AND TRIAL PLEASE(x^4-4x^3+3x^2+2x+6=0)I solved it using descarte's rule of sings...

You need to evaluate the derivative of the equation
of function such that:

`4x^3 - 12x^2
+ 6x+ 2 = 0`

You need to divide all equation
by 2 such that:

`2x^3 - 6x^2 + 3x+ 1
= 0`

You need to look the zeroes of derivative
among the following fractions: `+-1/1 ; +-1/2`
.

`x = -1 =gt -2 - 6 - 3 + 1 !=
0`

`x = 1 =gt 2 - 6 + 3
+ 1 = 0`

Notice that x = 1 cancels the
equation, hence you may write the equation of derivative such
that:

`2x^3 - 6x^2 + 3x + 1 = (x -
1)(ax^2 + bx+ c)`

`2x^3
- 6x^2 + 3x + 1 = ax^3 + bx^2 + cx - ax^2 - bx -
c`

Equating the coefficients of like powers
yields:

`a =
2`

`b -a = -6 =gt b - 2
= -6 =gt b = -4`

`c - b
= 3 =gt c + 4 = 3 =gt c = -1`

class="AM">`2x^3 - 6x^2 + 3x + 1 = (x - 1)(2x^2 - 4x -
1)`

You need to find the zeroes of class="AM">`2x^2 - 4x - 1 = 0`  such
that:

`x_(1,2) = (4 +- sqrt(16 +
8))/4`

`x_(1,2) = (4
+-2 sqrt6)/4`

`x_(1,2)
= (2 +- sqrt6)/2`

You need to evaluate the
value of function for each root of derivative such
that:

`x =1 =gt1-4+3+2+6 = 8
` ( + sign)

`x =-1 =gt
1+4+3-2+6 =12 gt 0`

`x
= 2 =gt 16-32+12+4+6 gt
0`

Hence, over the real set, the
values of the function are positive, hence, it means that the graph is floating above x
axis and the equation has no real
roots.

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