The sum of two zeroes of cubic polynomial $2 x^{3} - 13 x^{2} + 17 x + 12$ can be

\frac{- 7}{2}

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Solution

The correct option is D 7
From hit and trial method, x = 3 is a zero of the cubic polynomial.

Putting x = 3 in $2 x^{3} - 13 x^{2} + 17 x + 12,$

$2 (3^{3}) - 13 (3^{2}) + 17 (3) + 12$

$= 2 (27) - (13) 9 + (17) 3 + 12$

= 54 - 117 + 51 + 12

= 117 - 117

= 0

Thus, (x – 3) is the factor of $2 x^{3} - 13 x^{2} + 17 x + 12.$

$2 x^{3} - 13 x^{2} + 17 x + 12 = (x - 3) (2 x^{2} - 7 x - 4)$

By splitting the middle term:

$2 x^{2} - 7 x - 4$

$= 2 x^{2} - 8 x + x - 4$

$= 2 x (x - 4) + 1 (x - 4)$

$= (x - 4) (2 x + 1)$

$\Rightarrow x = 4, \frac{- 1}{2} .$

Thus, the zeroes of the polynomials $2 x^{3} - 13 x^{2} + 17 x + 12 a r e 3, - \frac{1}{2} and 4.$

Considering the zeroes 3 and 4, we get their sum as 3 + 4 = 7.

Hence, the correct answer is option (d).

Suggest Corrections

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