Find Descartes' rule of signs for `6x^4-x^3+4x^2-x-2`
Solution:
Here `f(x)=6x^4-x^3+4x^2-x-2`
Method-1:
`f(x)=color{blue}{+}6x^4color{red}{-}x^3color{blue}{+}4x^2color{red}{-}xcolor{red}{-}2`
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)` = | | + 6x^4 | | - x^3 | | + 4x^2 | | - x | | - 2 |
| Sign = | | + | | - | | + | | - | | - |
| Sign change count = | | | + to -
Change-1 | | - to +
Change-2 | | + to -
Change-3 | | - to -
No Change | |
There are 3 sign changes, so there are 3 or 1 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}{+}6(-x)^4color{red}{-}(-x)^3color{blue}{+}4(-x)^2color{red}{-}(-x)color{red}{-}2`
`f(-x)=color{blue}{+}6x^4color{blue}{+}x^3color{blue}{+}4x^2color{blue}{+}xcolor{red}{-}2`
| `f(-x)` = | | + 6x^4 | | + x^3 | | + 4x^2 | | + x | | - 2 |
| Sign = | | + | | + | | + | | + | | - |
| Sign change count = | | | + to +
No Change | | + to +
No Change | | + to +
No Change | | + to -
Change-1 | |
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}{+}6x^4color{red}{-}x^3color{blue}{+}4x^2color{red}{-}xcolor{red}{-}2`
The coefficients are `6,-1,4,-1,-2`
There are 3 sign changes, so there are 3 or 1 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}{+}6(-x)^4color{red}{-}(-x)^3color{blue}{+}4(-x)^2color{red}{-}(-x)color{red}{-}2`
`f(-x)=color{blue}{+}6x^4color{blue}{+}x^3color{blue}{+}4x^2color{blue}{+}xcolor{red}{-}2`
The coefficients are `6,1,4,1,-2`
There is 1 sign change, so there is exactly 1 negative roots.
3. For non-real roots (Imaginary roots) :
minimum number of non-real roots = degree - (positive roots + negative roots)
The degree is 4.
The maximum number of positive roots is 3.
The maximum number of negative roots is 1.
Hence, the minimum number of non-real roots is `4-(3+1)=0`.
Descartes's rule of signs chart
| Number of positive real roots | Number of negative real roots | Number of imaginary roots |
| 3 | 1 | `4-(3+1)=0` |
| 1 | 1 | `4-(1+1)=2` |
This material is intended as a summary. Use your textbook for detail explanation.
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