Find Descartes' rule of signs for `8x^4+7x-6`
Solution:
Here `f(x)=8x^4+7x-6`
Method-1:
`f(x)=color{blue}{+}8x^4color{blue}{+}7xcolor{red}{-}6`
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)` = | | + 8x^4 | | + 7x | | - 6 |
| Sign = | | + | | + | | - |
| Sign change count = | | | + to +
No Change | | + to -
Change-1 | |
There is 1 sign change, so there is exactly 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}{+}8(-x)^4color{blue}{+}7(-x)color{red}{-}6`
`f(-x)=color{blue}{+}8x^4color{red}{-}7xcolor{red}{-}6`
| `f(-x)` = | | + 8x^4 | | - 7x | | - 6 |
| Sign = | | + | | - | | - |
| Sign change count = | | | + 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}{+}8x^4color{blue}{+}7xcolor{red}{-}6`
The coefficients are `8,7,-6`
There is 1 sign change, so there is exactly 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}{+}8(-x)^4color{blue}{+}7(-x)color{red}{-}6`
`f(-x)=color{blue}{+}8x^4color{red}{-}7xcolor{red}{-}6`
The coefficients are `8,-7,-6`
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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