Q. 34.2( 5 Votes )

# Which of the foll

Answer :

We need to do hit and trial to find root of a cubic equation.

If it is a root of equation, it must satisfy the equation.

So, let’s start with option A.

(-2)^{3} + 2(-2)^{2} - 5(-2)-6 = -8 + 8 + 10 - 6 = 4

Let’s try option B

(2)^{3} + 2(2)^{2} - 5(2)-6 = 8 + 8 – 10 - 6 = 0

Let’s try option C

(-3)^{3} + 2(-3)^{2} - 5(-3)-6 = -27 + 18 + 15 - 6 = 0

For option D

(3)^{3} + 2(3)^{2} - 5(3)-6 = 27 + 18 – 15 - 6 = 24

Hence Option B and C are correct

Verifying –

Factors of the given equation is (x-2)(x + 3)(x + 1) = x^{3} + 2x^{2} - 5x - 6.

Rate this question :

The zeroes of theRS Aggarwal & V Aggarwal - Mathematics

Zero of the polynRS Aggarwal & V Aggarwal - Mathematics

If 2 and -1/3 areRS Aggarwal & V Aggarwal - Mathematics

If (x + 1) is a fRS Aggarwal & V Aggarwal - Mathematics

If (x^{2}RS Aggarwal & V Aggarwal - Mathematics

The zeroes of theRS Aggarwal & V Aggarwal - Mathematics

The zeroes of theRS Aggarwal & V Aggarwal - Mathematics

If p(x) = x^{3}RS Aggarwal & V Aggarwal - Mathematics

If p(x) = (x^{}RS Aggarwal & V Aggarwal - Mathematics

The coefficient oRS Aggarwal & V Aggarwal - Mathematics