Home > Algebra calculators > Descartes' rule of signs example

Descartes' rule of signs example ( Enter your problem )
  1. Method & Example-1 `(x^5-x^4+3x^3+9x^2-x+5)`
  2. Example-2 `(x^3+3x^2+3x+1)`
  3. Example-3 `(8x^4+7x-6)`
  4. Example-4 `(x^5+5x^4+10x^3+10x^2+5x+1)`
  5. Example-5 `(6x^4-x^3+4x^2-x-2)`

3. Example-3 `(8x^4+7x-6)`
(Previous example)
5. Example-5 `(6x^4-x^3+4x^2-x-2)`
(Next example)

4. Example-4 `(x^5+5x^4+10x^3+10x^2+5x+1)`





Find Descartes' rule of signs for `x^5+5x^4+10x^3+10x^2+5x+1`

Solution:
Here `f(x)=x^5+5x^4+10x^3+10x^2+5x+1`

Method-1:
`f(x)=color{blue}{+}x^5color{blue}{+}5x^4color{blue}{+}10x^3color{blue}{+}10x^2color{blue}{+}5xcolor{blue}{+}1`

1. For positive real roots:
The number of positive real roots of f(x) is the same as the number of changes in sign of the coefficients of f(x) or less than this by an even number.


`f(x)` = + x^5+ 5x^4+ 10x^3+ 10x^2+ 5x+ 1
Sign = ++++++
Sign change count = + to +
No Change
+ to +
No Change
+ to +
No Change
+ to +
No Change
+ to +
No Change


There are no sign changes, so there are no positive roots.


2. For negative real roots:
The number of negative real roots of the f(x) is the same as the number of changes in sign of the coefficients of f(-x) or less than this by an even number.


To find the number of negative real roots, substitute x with -x
`f(-x)=color{blue}{+}(-x)^5color{blue}{+}5(-x)^4color{blue}{+}10(-x)^3color{blue}{+}10(-x)^2color{blue}{+}5(-x)color{blue}{+}1`

`f(-x)=color{red}{-}x^5color{blue}{+}5x^4color{red}{-}10x^3color{blue}{+}10x^2color{red}{-}5xcolor{blue}{+}1`

`f(-x)` = - x^5+ 5x^4- 10x^3+ 10x^2- 5x+ 1
Sign = -+-+-+
Sign change count = - to +
Change-1
+ to -
Change-2
- to +
Change-3
+ to -
Change-4
- to +
Change-5


There are 5 sign changes, so there are 5 or 3 or 1 negative roots.


Method-2 for solution:

1. For positive real roots:
The number of positive real roots of f(x) is the same as the number of changes in sign of the coefficients of f(x) or less than this by an even number.


`f(x)=color{blue}{+}x^5color{blue}{+}5x^4color{blue}{+}10x^3color{blue}{+}10x^2color{blue}{+}5xcolor{blue}{+}1`

The coefficients are `1,5,10,10,5,1`

There are no sign changes, so there are no positive roots.


2. For negative real roots:
The number of negative real roots of the f(x) is the same as the number of changes in sign of the coefficients of f(-x) or less than this by an even number.


To find the number of negative real roots, substitute x with -x

`f(-x)=color{blue}{+}(-x)^5color{blue}{+}5(-x)^4color{blue}{+}10(-x)^3color{blue}{+}10(-x)^2color{blue}{+}5(-x)color{blue}{+}1`

`f(-x)=color{red}{-}x^5color{blue}{+}5x^4color{red}{-}10x^3color{blue}{+}10x^2color{red}{-}5xcolor{blue}{+}1`

The coefficients are `-1,5,-10,10,-5,1`

There are 5 sign changes, so there are 5 or 3 or 1 negative roots.



This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then Submit Here



3. Example-3 `(8x^4+7x-6)`
(Previous example)
5. Example-5 `(6x^4-x^3+4x^2-x-2)`
(Next example)





Share this solution or page with your friends.


 
Copyright © 2024. All rights reserved. Terms, Privacy
 
 

.