Answer:
Option D is correct that is Three real roots, each with a different value.
Step-by-step explanation:
Given Cubic Equation,
x³ + 6x² + 11x + 6
We have to find nature of roots of the equation.
We Answer by finding roots of the equation.
Let p(x) = x³ + 6x² + 11x + 6
put x = -1
p(-1) = (-1)³ + 6(-1)² + 11(-1) + 6 = -1 + 6 - 11 + 6 = 0
So, x = -1 is first zero and ( x + 1 ) is first factor.
To find other zeroes we divide p(x) by ( x + 1 )
On division, we get quotient = x² + 5x + 6
Zeroes of obtained quadratic polynomial are roots of p(x)
x² + 5x + 6 = 0
x² + 2x + 3x + 6 = 0
x( x + 2 ) + 3 ( x + 2 ) = 0
( x + 2 ) ( x + 3 ) = 0
x = -2 & -3
Thus, Roots of given cubic equation are -1 , -2 , -3
Therefore, Option D is correct that is Three real roots, each with a different value.
Pic is attached.
