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How many roots of the equation $3x^{4} + 6x^{3} + x^{2} + 6x + 3 = 0$ are real ?
Given :
Given equation is $f(x)$ = $3x^{4} + 6x^{3} + x^{2} + 6x + 3 = 0$
To find :
We have to find the number of real roots of the given equation.
Solution :
According to Descartes rule of signs,
The number of positive real zeros in a polynomial function f(x) is the same or less than by an even numbers as the number of changes in the sign of the coefficients.
The number of negative real zeros of the f(x) is the same as the number of changes in sign of the coefficients of the terms of $f(-x)$ or less than this by an even number.
In f((x)) there are no sign changes.
$f(-x) = 3(-x)^{4}+6(-x)^{3}+(-x)^{2}+6(-x)+3 = 0$
$ = 3x^{4} - 6x^{3} + x^{2} - 6x + 3 = 0$
In $f(-x)$, there are four sign changes. This implies, there are 4 or 2 real roots.
Comparing $f(x)$ and $f(-x)$, there are two sign changes.
Therefore, the given equation has two real roots.
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