Question

x8+x7+x5+6x3-8x-2 how many real roots are possible?

Answer 1

Answer:

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Answer 2

Answer:

Maximum number of real roots are 4.

Step-by-step explanation:

Descartes' Rule of Signs will tell me how many and what kind of roots I can anticipate, but it won't tell me where the polynomial's zeroes are (I'll need to apply the Rational Roots Test and synthetic division, or draw a graph, to actually locate the roots).

Descartes' Sign Rule: A method of determining the maximum number of positive and negative real roots of a polynomial.

For positive roots, start with the sign of the coefficient of the lowest (or highest) power. Count the number of sign changes n as you proceed from the lowest to the highest power (ignoring powers which do not appear). Then n is the maximum number of positive roots. Furthermore, the number of allowable roots is n, n-2, n-4, ...

For negative roots, starting with a polynomial f(x), write a new polynomial f(-x) with the signs of all odd powers reversed, while leaving the signs of the even powers unchanged. Then proceed as before to count the number of sign changes n. Then n is the maximum number of negative roots.

Given: $x^{8}+x^{7} +x^{5}+6x^{3} -8x-2$

Find: How many real roots are possible?

Let $f(x)=x^{8}+x^{7} +x^{5}+6x^{3} -8x-2$

                     $++++--$

Here, only one sign changes, so maximum one possible positive roots.

Now, $f(-x)=x^{8}-x^{7} -x^{5}-6x^{3} +8x-2$

$+---+-$

There are three sign changes, so there are a maximum of three negative roots.

Hence, maximum number of real roots $=1+3=4$

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