Question

"If all the zeroes of a cubic polynomial $x^3+a{x}^2-bx+c$ are position, then atleast one of $a,b$ and $c$ is one-negative." Is this statement true of false? Justify.

Answer 1

The statement is false.

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

$\alpha \beta \gamma ={(-1)}^3\frac{constantterm}{Coefficientof{x}^3}=\frac{-c}{1}$

$\therefore \alpha \beta \gamma =-c$

Given that, all three zeroes are positive.

So, the product of all three zeroes is also positive.

$\Rightarrow \alpha \beta \gamma >0$

$\Rightarrow -c>0$

$\Rightarrow c<0$

Now, sum of the zeroes$=\alpha +\beta +\gamma =(-1)\frac{coefficientof{x}^2}{Coefficientof{x}^3}=\frac{-a}{1}=-a$

But $\alpha ,\beta$ and $\gamma$ are all positive.

Thus, its sum is also positive.

$\Rightarrow \alpha +\beta +\gamma >0$

$\Rightarrow -a>0$

$\Rightarrow a<0$

And sum of the product of two zeroes at times $={(-1)}^2\frac{coefficientofx}{Coefficientof{x}^3}=\frac{-b}{1}=-b$

$\alpha \beta +\beta \gamma +\gamma \alpha =-b>0$

$\Rightarrow -b>0$

$\Rightarrow b<0$

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

So, the cubic polynomial has all three zeroes which are positive only when all constants $a,b$ and $c$ are negative.