how do i determine the number of real zeros of the function
X^3 + 4 X^2 + X -6
Sum of the roots has to be : X1 + X2 + X3 = -4
X1 . X2. X3 = -6 write down all factors of 6 : 1,2, 3
we see that -3 - 2 +1 =-4
and (-3)( -2) (1) = -6
Therefore the 3 rational roots are: X1 = -3 X2 = -2 X3=1
Polynomial can be factored as: X^3 + 4 X^2 + X - 6 = ( X+3) ( X+2) ( X-1)
This is what you do without calculation:
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