Using descartes' rule of signs to describe the roots of h (x) = 4x^4-5x^3+2x^2-x+5, how many total roots must there be in this fourth-degree function?

Question

Mathematics

Jonathon Hogan

12 July, 08:02

Using descartes' rule of signs to describe the roots of h (x) = 4x^4-5x^3+2x^2-x+5, how many total roots must there be in this fourth-degree function?

+1

Answers (1)

Know the Answer?

Yesenia Goodwin · Answer 1

H (x) = 4x⁴ - 5x³ + 2x² - x + 5

Take the coefficients, we have:

4 - 5 2 - 1 5

From the first coefficient to the second coefficient, there's a change of sign from positive to negative (positive four to negative five)

From the second coefficient to the third coefficient, there's a change of sign from negative to positive (negative five to positive two)

From the third coefficient to the fourth coefficient, there's a change of sign from positive to negative (positive two to negative one)

From the fourth coefficient to the fifth coefficient, there's a change of sign from negative to positive (negative one to positive five)

The changing of sign happened four times, and it means h (x) has four positive real zeros or less (even numbers of zeros), so h (x) could have either four or two real zeros.

The maximum number of solutions for a polynomial of degree four is four solution, so there are no negative zeros.