Upper and Lower Bounds of Zeros of Polynomials

 

The What (what you need to know)
    A positive real number is an upper bound on the zeros of a polynomial (meaning there are no real zeros larger than it), if when you divide that polynomial by x minus that number synthetically the results line including the remainder all have the same sign. If you get any zeros, they act like wild cards in that they can count as a positive or as a negative.

    A negative real number is a lower bound on the zeros of a polynomial (meaning there are no real zeros smaller than it), if when you divide that polynomial by x minus that number synthetically the results line including the remainder alternate in sign. Again, any zeros you get can count either way.

The How (how you apply it)
    Example: Let f(x)= -3x4 + 5x3 + 5x2 + 19x+6. Show that 3 is an upper bound of the zeros and -1 is a lower bound of the zeros.   When you divide f(x) by x - 3 synthetically, you get -3  -4  -7  -2   0. These are all the same sign, so 3 is an upper bound on the real zeros.   When you divide f(x) by x + 1 synthetically, you get -3  8  -3  22   -16. These signs alternate, so -1 is a lower bound.

The Why (why it works)
    It is easy to see why these work, if you rewrite f(x) in terms of what you got after the division: 
          

                         f(x) = -3x4 + 5x3 + 5x2 + 19x + 6 = (x – 3)(-3x3 – 4x2 – 7x – 2) + 0
   Now, to show why same signs indicate an upper bound, consider what happens when you plug in a number larger than three: x – 3 is positive, -3x3 – 4x2 – 7x – 2 is negative, since every term is negative.   Thus (x – 3)(-3x3 – 4x2 – 7x – 2) = positive × negative = negative.  The rule says that the remainder must be the same sign also: a negative plus a negative = a negative.  Zero counts either way, so in this case a negative plus zero = a negative.  The point is that since the result must be negative in this case, then it can’t be zero, so we won’t get another zero.

                    f(x) = -3x4 + 5x3 + 5x2 + 19x + 6 = (x + 1)(-3x3 + 8x2 – 3x + 22) – 16
    Next, we show why alternating signs indicate lower bound. Consider what happens when you plug in a number less than negative one: x + 1 is negative, -3x3 is positive, 8x2 is positive, -3x is positive, 22 is positive. Thus -3x3 + 8x2 – 3x + 22 is positive. Hence (x + 1)(-3x3 + 8x2 – 3x + 22) = negative × positive = negative. Add –16 to that negative and the result is still negative. Again the result is definitely negative, so it can’t be zero.

The Practice (problems for you to try)

1.)   Use the Rational Root Theorem and the property of upper bounds and lower bounds of zeros of polynomials to find the guaranteed smallest upper bound of the possible rational zeros and the guaranteed largest lower bound of the possible rational zeros on the following function.

g(x) = 3x4 + 7x3 – 21x2 – 35x + 30

2.)   You want to find the real zeros of a function k(x) graphically. You find that when you divide k(x)

by: you get:
x - 7 8x4 + 12x3 + 118x2 + 897x + 6228 with remainder 43560
x - 6 8x4 + 4x3 + 58x2 + 419x + 2463 with remainder 14742
x - 18x4 - 36x3 - 2x2 + 69x + 18 with remainder -18
x + 4 8x4 - 76x3 + 338x2 - 1281x + 5073 with remainder -20328
x + 3 8x4 - 68x3 + 238x2 - 643x + 1878 with remainder -5670
x + 28x4 - 60x3 + 154x2 - 237x + 423 with remainder -882
x + 18x4 - 52x3 + 86x2 - 15x - 36 with remainder 0
Using only this information and letting Ymin = -1 and Ymax = 1, what is the smallest viewing rectangle that should be used to find all the real zeros of k(x) graphically in your calculator? Xmin = _____________, Xmax = ____________

 

Answers:

1.)   The possible rational roots are ± 1, ± 2, ± 3, ± 5, ± 6, ± 10, ± 15, ± 30, ± 1/3, ± 2/3, ± 5/3, ± 10/3.

First, plug in positive numbers to find the smallest possible upper bound of the possible rational zeros:

3 is the smallest upper bound of the possible rational zeros. We know this since it is the smallest of the possible rational zeros that results in all the same sign in the results line of the synthetic division.

Next, plug in the negative numbers to find the lower bound:

-5 is the largest guaranteed lower bound on the real zeros of the function due to the property of lower bounds. If you graph the function using the window xmin = -5 and xmax = 3 you will find that -3 is a lower bound of the real zeros, but you can’t tell that from the synthetic division.

2.)   Xmin = ____-2_________, Xmax = _____6_______