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:

+ 6x+ 2 = 0

You need to divide all equation
by 2 such that:

= 0

You need to look the zeroes of derivative
among the following fractions:
.

0

+ 1 = 0

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

1)(ax^2 + bx+ c)

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

Equating the coefficients of like powers
yields:

2

= -6 =gt b = -4

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

class="AM">
1)

You need to find the zeroes of class="AM">  such
that:

8))/4

+-2 sqrt6)/4

= (2 +- sqrt6)/2

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

( + sign)

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

= 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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