Question

find the zeros of P(x)= 6x⁴-7x³+56x²-63x+18. include descartes rule of signs,upper bound, etc. thanks

Answer 1

The function is

Apply Descartes rule of signs, from the polynomial find the number of sign changes.

Positive Root Case:

There are four sign changes in P(x) from + to -, - to +, + to - and - to + implies there are maximum 0,2,4 positive roots

Negative Root Case:

That is

There are NO sign changes in P(x) implies there are NO negative roots

---

From Rational zeroes, roots are of the form where p is factor of constant term and q is factor of the leading coefficient. Factors of 18 are 1,2,3,6,9,18 and factors of 6 are 1,2,3,6

Since there are no negative roots implies possible roots are

Find where f(root)=0 implies

Hence, there are 2 positive roots implies are factors of P(x). Divide polynomial P(x) by the above two factors using long division

Hence, P(x) can be written as

Roots are