Arranging the real number in ascending order -8,2,0,9,-10
2 answers:
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Solution:
There is no (zero) complex zeros for the given polynomial.
Explanation:
We apply Descartes' rule of sign to identify the number of complex roots.
The given polynomial is
Let us see the number of sign changes in f(x)
+,-,+
There are 2 sign changes in f(x). One from plus to minus and second from plus to minus. Hence, there 2 or 0 positive roots.
Now, let us see the number of sign changes in f(-x)
-,-,+
There are only one sign change. Hence, there will be 1 negative roots.
The degree of the polynomial is 3. Hence, there will be exactly 3 zeros.
Therefore, the possible numbers of zeros are
2 positive, 1 negative and 0 complex
0 positive, 1 negative and 2 complex.
Hence, possible number of complex zeros are 0 and 2.
Now, we graph the function in xy plane and see that the graph cuts the x axis at three points. It means all the zeros are real , which is the case of first possibility (2 positive, 1 negative and 0 complex).
Hence, the number of complex zeros for the given polynomial is zero.
