Question

If a, b and c are the zeroes of a cubic polynomial ${\mathrm{px}}^3+{\mathrm{qx}}^2+\mathrm{rx}+s\mathrm{and}0>a>b>c,$ then which of the following is true?

• A. p > 0, q > 0, r > 0, s > 0
• B. p < 0, q < 0, r < 0, s > 0
• C. p > 0, q > 0, r > 0, s < 0
• D. p > 0, q < 0, r < 0, s > 0

Answer 1

The correct option is A p > 0, q > 0, r > 0, s > 0

Given that a, b, c are the zeroes of ${\mathrm{px}}^3+{\mathrm{qx}}^2+\mathrm{rx}+s\mathrm{such}\mathrm{that}0>a>b>c.$

⇒ All zeroes are negative

$\therefore a+b+c=\frac{-q}{p}<0$ (∵ Sum of negative numbers is negative)

$\Rightarrow q>0,p>0\mathrm{or}q<0,p<0$ ...(i)
$\mathrm{Also},\mathrm{ab}+\mathrm{bc}+\mathrm{ca}=\frac{r}{p}>0$ (∵ Product of 2 negative numbers is positive and sum of positive numbers is positive)

$\Rightarrow r>0,p>0\mathrm{or}r<0,p<0$ ...(ii)

$\mathrm{Also},\mathrm{abc}=\frac{-s}{p}<0$ (∵ Product of 3 negative numbers is negative)

$\Rightarrow s>0,p>0\mathrm{or}s<0,p<0$ ...(iii)

From (i), (ii) and (iii),

$p>0,q>0,r>0,s>0\mathrm{or}p<0,q<0,r<0,s<0.$

Hence, the correct answer is option (1).