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Find the cubic polynomial with the sum of zeroes, sum of product of its zeroes taken 2 at a time and product of its zeroes as 0, -7 and -6 respectively.

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x^{3} - 7 x - 6

x^{3} - 7 x^{2} + 6

x^{3} - 7 x + 6

x^{3} - 7 x^{2} - 6

answer is C.

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Detailed Solution

We are given that
α+β+γ = −ba = 0
αβ+βγ+αγ=ca = −7
αβγ = −da = −6
Where a, b, c are the coefficients of

x^{3}, x^{2}, x

respectively and d is the constant part.
So we will substitute the values and get the final equations as

\Rightarrow - \frac{b}{a} = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (i)

\Rightarrow \frac{c}{a} = - 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (ii)

\Rightarrow - \frac{d}{a} = - 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (iii)

On solving equation (i) we will get

\Rightarrow - \frac{b}{a} = 0

\Rightarrow - b = 0

\Rightarrow b = 0

From equation (ii) we will get

\Rightarrow \frac{c}{a} = - 7

\Rightarrow c = - 7 a

From equation (iii) we will get

\Rightarrow - \frac{d}{a} = - 6

\Rightarrow \frac{d}{a} = 6

\Rightarrow d = 6 a

Now we know that the general cubic polynomial is of the form

a x^{3} + b x^{2} + cx + d = 0

So let us put the values of b, c and d as we get it from the above equations.

∴ a x^{3} + b x^{2} + cx + d = 0

\Rightarrow a x^{3} + 0 \times x^{2} + (- 7 a) x + 6 a = 0

\Rightarrow a x^{3} - 7 ax + 6 a = 0

\Rightarrow a (x^{3} - 7 x + 6) = 0

\Rightarrow x^{3} - 7 x + 6 = 0

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