Using Descartes' Rule of Signs

Nov 2008
62
2
Use Descartes' rule of signs to determine the possible number of:
a) positive real roots of \(\displaystyle 4x^3 + 2x^2 + 7x + 9 = 0\)
b) negative real roots of \(\displaystyle 4x^3 + 2x^2 + 7x + 9 = 0\)


My answer to a) (zero positive real roots), was correct.

Zero positive real roots because there are zero sign changes..

and I had thought there would be zero negative roots as well. But it came back wrong.

Can someone explain to me why? Thanks.
 

masters

MHF Helper
Jan 2008
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Big Stone Gap, Virginia
Use Descartes' rule of signs to determine the possible number of:
a) positive real roots of \(\displaystyle 4x^3 + 2x^2 + 7x + 9 = 0\)
b) negative real roots of \(\displaystyle 4x^3 + 2x^2 + 7x + 9 = 0\)


My answer to a) (zero positive real roots), was correct.

Zero positive real roots because there are zero sign changes..

and I had thought there would be zero negative roots as well. But it came back wrong.

Can someone explain to me why? Thanks.
Hi Savior_Self,

\(\displaystyle f(-x)=4(-x)^3+2(-x)^2+7(-x)+9\)

\(\displaystyle f(-x)=-4x^3+2x^2-7x+9\)

Now, how many sign changes do you see?
 
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May 2010
43
7
To find the maximum number of positive real roots count the number of sign changes. To find the number of negative real roots replace all x values with a (-x) and perform the same operation of counting sign changes.

For example,

\(\displaystyle x^2+x+2\) becomes...
\(\displaystyle (-x)^2+(-x)+2\)
 
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Jun 2008
1,389
513
Illinois
Using Descarte's method, to determine the number of negative roots you replace the x in the polynomial with -x, and see how many sign changes there are in the resulting equation:

\(\displaystyle
4(-x)^3 + 2(-x)^2 + 7(-x) + 9 =
\)
\(\displaystyle
-4x^3 + 2x^2 - 7x + 9
\)

Note that there are 3 changes of sign as you read left to right. Thus there may be 3 negative roots, or possibly just 1 (since it's possible that 2 of the roots may actually be complex).

If you plot this function you'll see that there is one real negtive root. The two other roots are therefore complex.
 
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Nov 2008
62
2
Awesome. Thanks guys. I didn't pick up on the x to (-x) step for some reason. I won't forget it now.
 
Jun 2008
1,389
513
Illinois
There's another "trick" that you could have used. Remember that a polynomial of order N must have N roots. So since you already knew that there were zero positive roots, you know that there must be either (a) 3 negative roots, or (b) 1 negative root and 2 complex roots.