F(x) = 4x^4 – 2x^3 – 3x^2 + 6x - 9 Find the Zeros Using Descartes' Rule of Signs

So possibilities of zeroes are:

Positive       Negative       Imaginary

   1                    1                   2

   3                   1                   0

Zeroes = -1.4549, 1.2658, 0.34457-1.0503i, 0.34457+1.0503i.  

Step-by-step explanation:

Note: Descartes' Rule of Signs is used to find the signs of zeroes not the exact value.

The given function is

Degree of polynomial is 4 so number of zeroes is 4.

There are three sign changes, so there are either 3 positive zeros or 1 positive zero.

Now, put x=-x in f(x).

There is one variation in sign change, so there is 1 negative zero.

So possibilities of zeroes are:

Positive       Negative       Imaginary

   1                    1                   2

   3                   1                   0

Using graphing calculator the zeroes of given function are -1.4549, 1.2658, 0.34457-1.0503i and 0.34457+1.0503i.  

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step-by-step explanation:

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