Descartes' rule of signs Method & Example-1 (x^5-x^4+3x^3+9x^2-x+5)

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Descartes' rule of signs example ( Enter your problem )

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Method & Example-1 `(x^5-x^4+3x^3+9x^2-x+5)`
Example-2 `(x^3+3x^2+3x+1)`
Example-3 `(8x^4+7x-6)`
Example-4 `(x^5+5x^4+10x^3+10x^2+5x+1)`
Example-5 `(6x^4-x^3+4x^2-x-2)`

2. Example-2 `(x^3+3x^2+3x+1)`
(Next example)

1. Method & Example-1 `(x^5-x^4+3x^3+9x^2-x+5)`

Descartes' Rule of Signs :
Help us to find out positive real roots and negative real roots of a polynomial `f(x)`

Explanation of Descartes' Rule of Signs :
1) `f(x)` is a polynomial where the exponents are arranged from highest to lowest, with real coefficients excluding zero.
2) If `n` is the number of sign changes in `f(x)`, then the number of positive real roots is either `n` or `n-2` or `n-4` etc.
3) If `n` is the number of sign changes in `f(-x)`, then the number of negative real roots is either `n` or `n-2` or `n-4` etc.

Examples for both cases :
If 5 is the number of sign changes in `f(x)`, then the number of positive real roots is either 5 or 3 or 1
If 6 is the number of sign changes in `f(-x)`, then the number of negative real roots is either 6 or 4 or 2 or 0

If two signs of adjacent coefficients
1) are alternate, then it is considered as sign change (eg: + to - or - to +)
2) are same, then it is considered as no sign change (eg: + to + or - to -)

Example-1
Find Descartes' rule of signs for `x^5-x^4+3x^3+9x^2-x+5`

Solution:
Here `f(x)=x^5-x^4+3x^3+9x^2-x+5`

Method-1:
`f(x)=color{blue}{+}x^5color{red}{-}x^4color{blue}{+}3x^3color{blue}{+}9x^2color{red}{-}xcolor{blue}{+}5`

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

- x^4

+ 3x^3

+ 9x^2

- x

+ 5

Sign =

Sign change count =

+ to -
Change-1

- to +
Change-2

+ to +
No Change

+ to -
Change-3

- to +
Change-4

There are 4 sign changes, so there are 4 or 2 or 0 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{red}{-}(-x)^4color{blue}{+}3(-x)^3color{blue}{+}9(-x)^2color{red}{-}(-x)color{blue}{+}5`

`f(-x)=color{red}{-}x^5color{red}{-}x^4color{red}{-}3x^3color{blue}{+}9x^2color{blue}{+}xcolor{blue}{+}5`

`f(-x)` =

- x^5

- x^4

- 3x^3

+ 9x^2

+ x

+ 5

Sign =

Sign change count =

- to -
No Change

- to +
Change-1

+ to +
No Change

There is 1 sign change, so there is exactly 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{red}{-}x^4color{blue}{+}3x^3color{blue}{+}9x^2color{red}{-}xcolor{blue}{+}5`

The coefficients are `1,-1,3,9,-1,5`

There are 4 sign changes, so there are 4 or 2 or 0 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{red}{-}(-x)^4color{blue}{+}3(-x)^3color{blue}{+}9(-x)^2color{red}{-}(-x)color{blue}{+}5`

`f(-x)=color{red}{-}x^5color{red}{-}x^4color{red}{-}3x^3color{blue}{+}9x^2color{blue}{+}xcolor{blue}{+}5`

The coefficients are `-1,-1,-3,9,1,5`

There is 1 sign change, so there is exactly 1 negative roots.

This material is intended as a summary. Use your textbook for detail explanation.
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2. Example-2 `(x^3+3x^2+3x+1)`
(Next example)

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