Answer to Question #324540 in Algebra for Busi

Question #324540

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.


1
Expert's answer
2022-04-08T13:28:28-0400

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.

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