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?

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. 

I’m going to infer you said parallel.solve for yy= -1/2x+7/4lines that are parallel have the same slope, different y intercept.so we know m is -1/2then from there solve for y intercept.