Zeros of Polynomials - Finding open interval

• Mar 13th 2010, 10:52 PM
tottijohn
Zeros of Polynomials - Finding open interval
Hi, I have a question for the above topic.

Let p(x) =5x^3 -23x^2 + x + 6 be a polynomial. Then all zeros of p(x) lies in [-M,M]. Find the value of M to 2 d.p.

My solution
p(x) = x^3 - (23/5)x^2 + (1/5)x +6/5
M = 1 + max{|-23/5|, |1/5|,|6/5|}
= 1 + 23/5 = 5.6

Here, i am not sure if i should divide the constant 6 by the leading coefficient which is 5. If i don't divide, then M = 1 + 6 = 7.

So, which is the correct one?
• Mar 14th 2010, 12:22 AM
CaptainBlack
Quote:

Originally Posted by tottijohn
Hi, I have a question for the above topic.

Let p(x) =5x^3 -23x^2 + x + 6 be a polynomial. Then all zeros of p(x) lies in [-M,M]. Find the value of M to 2 d.p.

My solution
p(x) = x^3 - (23/5)x^2 + (1/5)x +6/5
M = 1 + max{|-23/5|, |1/5|,|6/5|}
= 1 + 23/5 = 5.6

Here, i am not sure if i should divide the constant 6 by the leading coefficient which is 5. If i don't divide, then M = 1 + 6 = 7.

So, which is the correct one?

What you have done is a correct application of the Cauchy bound for the roots of a polynomial, that is it gives 5.6.

In fact the largest absolute value of a root is about 4.5

CB
• Mar 14th 2010, 03:13 AM
tottijohn
Thank you, CaptainBlack for the prompt reply!