Q. 7 A

Find the zeroes o

Answer :

For finding zeroes, we’ll use the hit and trial method.

For that, we’ll put a value which will make our polynomial zero.


Since, here the whole equation is positive that means, zero can be achieved by putting the negative values only.


So, first we’ll put x = – 1.


(– 1)3 + 6(– 1)2 + 11(– 1) + 6 = – 1 + 6 – 11 + 6


= 0


x = – 1 is one of the zeroes.


Now, we’ll write the equation in terms of (x + 1) to make equation quadratic and hence to find the other zeroes.


x3 + 6x2 + 11x + 6 = x2 (x + 1) + 5x(x + 1) + 6(x + 1)


= (x + 1)(x2 + 5x + 6)


Now, using factorization by splitting the middle term method


= (x + 1)(x2 + (3 + 2)x + 6)


= (x + 1)(x2 + 3x + 2x + 6)


= (x + 1)(x(x + 3) + 2(x + 3))


= (x + 1)(x + 2)(x + 3)


Now, to get the zeroes we’ll put these equals to zero.


x + 1 = 0, x + 2 = 0, x + 3 = 0


x = – 1, x = – 2, x = – 3 are the zeroes of x3 + 6x2 + 11x + 6


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