Consider the following function.

Question

f(x) = -7x³ + 44x² - 68x + 16

(a) Use Descartes's Rule of Signs to determine the possible numbers of positive real zeros of f. (Select all that apply.)

Select: 0,1,2,3,4,5

Use Descartes's Rule of Signs to determine the possible numbers of negative real zeros of f. (Select all that apply.)

Select: 0,1,2,3,4,5

(b) Use the Rational Zero Test to create a list of all the possible rational zeros of f. (Enter your answers as a comma-separated list.)

x =

(c) Use a graphing utility to graph f so that some of the possible zeros in parts (a) and (b) can be disregarded.

(d) Determine all the real zeros of f. (Enter your answers as a comma-separated list.)

Answer 1

Nice problem. There are 3 sign reversals for a polynomial of degree 3, so there can be up to three positive roots and no negative roots. As complex roots must be in pairs, there are either 1 or 3 positive roots.

b) Since the leading coefficient is 7 and the constant coefficient is 16, roots can be

1/7, 2/7, 4/7, 8/7, 16/7, 1, 2, 4, 8, 16

If the roots must be of increasing order, they are

1/7, 2/7, 4/7, 1, 8/7, 2, 16/7, 4, 8, 16

c) I invite you to graph it.

You will see that you have integer roots at 2 and 4

As the product of the roots is 16/7, the third root is 2/7. (The graphing utility would eliminate all of the others. You could argue that 2/7 is not seen on the graph, but clearly it works; you could argue if you wished that the roots are 2, 4, and between .28 and .29, so the other 7 rational roots are eliminated, and the only other possible root is 2/7, so you check this (or use the factor that the product of roots is 16/7, as mentioned above).

d) 2/7, 2, 4

Not asked, but the polynomial would thus be factored as (7x - 2)(x-2)(x-4)