Question

If all three zeroes of a cubic polynomial x³ + ax² – bx + c are positive, then at least one of a, b and c is non-negative.

Options

• True

• False

Answer 1

This statement is False.

Explanation:

Let α, β and γ be the three zeroes of cubic polynomial x³ + ax² – bx + c.

Then, product of zeroes = `(-("Constant  term"))/("Coefficient of"  x^3)`

`\implies` αβγ = `c/1`

`\implies` αβγ = `-c`   ......(i)

Given that, all three zeroes are positive.

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

i.e., αβγ > 0

`\implies` – c > 0  .....[From (i)]

`\implies` c < 0

Now, sum of the zeroes = α + β + γ

= `(-("Coefficient of"  x^2))/("Coefficient of"  x^3)`

`\implies` α + β + γ = `a/1 = -a`

But α, β and γ all are positive.

Thus, their sum is also positive.

So, α + β + γ > 0

`\implies`  – a > 0

`\implies` a < 0

And sum of the product of the zeroes taken two at a time

= `("Coefficient of"  x)/("Coefficient of"  x^3)`

= `(-b)/1`

`\implies` αβ + βγ + γα = `- b`

∵ αβ + βγ + αγ > 0

`\implies` `-b > 0`

`\implies` b < 0

∴  All the coefficients a, b and c are negative.