Possible Roots?

Question

P(x) = x5 - 4x4 + 3x3 + 2x – 6
Possible Roots:   

 

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

 

 

Answer 1

Hi Lizzie.

 

Descarte's rule of signs states that for a polynomial who's terms are ordered in decreasing powers, the number of possible roots will either be the number of sign changes in the term, or a value less than the number of sign changes by an even number.

 

If we look at your function:

 

P(x) = +x^5 - 4x^4 + 3x^3 + 2x - 6

 

there are a total of 3 sign changes. Positive to negative, negative to positive, then positive to negative again. That means there will either be 3 roots or 1 root.

 

For higher-powered functions like this, it's best to try out some test points to see if you can get a zero from it. If a test point winds up with zero, you know it's a factor, so you can use long division to bring the power of the whole function down to easier levels.

 

Lets try x=1

 

P(1) = 1 - 4 + 3 + 2 - 6 = -4

 

That didn't work.

 

P(-1) = -1 - 4 -3 - 2 - 6 = -16

 

That didn't work either.

 

P(2) = 32 - 64 + 24 + 4 - 6 = -10

 

Still no luck.

 

P(3) = 243 - 324 + 81 + 6 - 6 = 0

 

Finally.

 

Since 3 is a zero you know that (x-3) is a factor. So use long division to bring the power down.

 

(x^5 - 4x^4 + 3x^3 + 2x - 6) / (x - 3) = x^4 - x^3 + 2

 

So now you have to look for zeros of the function:

 

f(x) = x^4 - x^3 + 2

 

This function will always be greater than 0. So there aren't any roots from it.

 

Your only possible root for P(x) then is x = 3, which is perfectly allowed by Descarte's rule.

 

Hope this helps.