Use the theorem on bounds to establish the best integral bounds for the roots of the following equation. 5x^3-17x^2+144=0

Answer 1

I'm going to arbitrarily pick -5 as a place to check for lower bounds. Doing a synthetic division we get:

This has the +/-/+/- characteristics that tell me there are no zero's below this, however, I don't know if this is THE lower bound (there could be one higher than this).

Lets try -3

Also has the +/-/+/- characteristics that also tells me there are no zero's below this, however, I still don't know if this is THE lower bound (there could be one higher than this).

Let's try -2

Notice this does not have the +/-/+/- characteristic. Also, the remainder is positive, meaning f(-2) = 36. Therefore, there must be a zero between -2 and -3.

Now we also know that -3 is the Integral Lower Bound

The Upper Bound will show the characteristic of all positive numbers in the synthetic division result. To get all positives, I need to overcome the 2nd coefficient -17 and since the first coefficient is 5, I need to multiply be 4 (yielding 20) to get a positive when I add it to -17, so let's try x = 4

So x = 4 is my Integral Upper Bound

So my Integral Bounds for the zeros are (-3, 4). Any and all zeros will be between those integers.

Answer 2

see http://www.mesacc.edu/~marfv02121/readings/bounds/ for the theorem

Clearly 3 is insufficent, as 5*3 = 15, and -17 + 15 < 0

Let's try 4.

4|5 -17 0 144

20 12 48

5 3 12 192

4 is an upper bound using the theorem (signs stay the same)

-3|5 -17 0 144

-15 96 -288

5 -32 96 -144

-3 is a lower bound (signs alternate)

To show that -2 does not work

-2|5 -17 0 144

-10 54 -108

5 -27 54 36 The last two signs don't alternate.

Note that a plot can show that there is only one root, and it is between -3 and -2.