Answer to Question #324540 in Algebra for Busi

Question

Use the Upper and Lower Bounds Theorem to show that the real zeros of

(i) P (x) = 7x^8-2x^5+x^2-2 lie between −1 and 1.

Answer 1

Upper Bound

If you divide a polynomial function f(x) by (x - c), where c > 0, using synthetic division and this yields all positive numbers, then c is an upper bound to the real roots of the equation f(x) = 0. An upper bound is an integer greater than or equal to the greatest real zero.

Lower Bound

If you divide a polynomial function f(x) by (x - c), where c < 0, using synthetic division and this yields alternating signs, then c is a lower bound to the real roots of the equation f(x) = 0. A lower bound is an integer less than or equal to the least real zero.

Task

Show that the real zeros of P(x) = 7x⁸-2x⁵+x²-2 lie between −1 and 1.

In other words, we need to show that -1 is a lower bound and 1 is an upper bound for real roots of the given equation.

Checking the Lower Bound:

Lets apply synthetic division with -1 and see if we get alternating signs. (see pic.1)

The values are of the alternating sign. Hence, -1 is the lower boundary.

Checking the Upper Bound:

Lets apply synthetic division with 1 and see if we get all positive. (see pic.2)

All the coefficients and the remainder are positive. So 1 is the upper boundary.

Since -1 is a lower bound and 1 is an upper bound for the real roots of the equation, then that means all real roots of the equation lie between -1 and 1.