Possible Roots?

Question

P(x) = -5x4 + x3 + 2x2 – 1

 

Use Descartes' Rule of Signs to analyze the zeros of the following. List all possibilities.

 

Please explain

Answer 1

Hi Lizzie,

 

Descates' rule of signs says if we have a single variable and order the equation in descending order of exponents, we then look at the sign changes.

 

In P(x) = -5x⁴ + x³ + 2x² -1

 

Here we have a change from -5x⁴ to x³ and a change from 2x² to -1

 

Therefore there are two possible positive roots.

 

To test for negative roots...we rewrite the polynomial as P(-x), changing the signs of the odd exponents and keeping the signs of the even exponents.

 

P(-x) = -5x⁴ - x³ + 2x² - 1

Then count the sign changes.

Again we have 2 changes

-x³ to 2x²

2x² to -1

which gives us 2 negative roots.

 

We would expect a maximum of 2 positive roots and 2 negative roots.

Answer 2

Rene Descartes was a merry old fart but his rules will help you solve.

 

For the positive zeroes, look at your equation and note how many times the sign changes

 

-5x⁴ + x³ + 2x² - 1  

 

You go from negative to positive stay at positive and then back to negative.  

This means you might have 2 positive solutions (zeroes), but you always have the possibility that these two may be complex (imaginary numbers) so you subtract 2 and find out you MIGHT have 0 positive roots so both are possibilities.

 

If you had counted eight sign changes, then there could be 8 positive roots, but there may be 6, or 4, or 2, or 0.

 

Now you want to count your negative zeroes.  To do this you want to set it up to be P(-x).

 

This gives you

-5(-x)⁴ + (-x)³ + 2(-x)² - 1 =

-5x4 - x3 + 2x2 - 1   <---- a shortcut is to simply change the signs on the terms with an odd exponent like the case of x³ in your problem

 

Lastly, count the number of sign changes in this new problem and you see you go from

negative to negative to positive to negative which gives you 2 possible negative roots, but like above, it is also possible that you have 0 negative roots, both are options.