Are the following statements 'True' or 'False'? Justify your answers.
If all three zeroes of a cubic polynomial \( x^{3}+a x^{2}-b x+c \) are positive, then at least one of \( a, b \) and \( c \) is non-negative.

Given:

If all three zeroes of a cubic polynomial \( x^{3}+a x^{2}-b x+c \) are positive, then at least one of \( a, b \) and \( c \) is non-negative.

To do:

We have to find whether the given statement is true or false.

Solution:

Let $\alpha, \beta$ and $ \gamma$ be the zeroes of cubic polynomial $x^{3}+a x^{2}-b x+c$

This implies,

Product of zeroes $=\alpha \beta \gamma=-\frac{\text { constant term }}{\text { Coefficient of } \mathrm{x}^{3}}$

$=\frac{-c}{1}$

$\alpha \beta \gamma=-c$

Given that, all three zeroes are positive.

This implies,

The product of all three zeroes is also positive.

$\alpha \beta \gamma>0$

$-c>0$

$c<0$

Sum of the zeroes $=\alpha+\beta+\gamma=-\frac{\text { coefficient of } \mathrm{x}^{2}}{\text { Coefficient of } \mathrm{x}^{3}}$

$=\frac{-a}{1}$

$=-a$

But $\alpha, \beta$ and $\gamma$ are all positive.
This implies, its sum is also positive.
$\alpha+\beta+\gamma>0$

$-a>0$

$a<0$

Sum of the product of two zeroes at a time $=\frac{\text { coefficient of } \mathrm{x}}{\text { Coefficient of } \mathrm{x}^{3}}$

$=\frac{-b}{1}$

$= -b$
Therefore, the cubic polynomial $x^{3}+a x^{2}-b x+c$ has all three zeroes which are positive only when all constants $a, b$ and $c$ are negative.

Hence, the given statement is false.

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Updated on: 10-Oct-2022

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