Asked by sanjay
If one of the zeroes of the cubic polynomial x3+ax2+bx+c is -1 then the product of the other two zeroes is -------(A) b -a +1 (B) b-a-1 (C) a-b+1 (D) a-b-1
Answers
Answered by
Steve
The product of all three roots is -c. So, if one root is -1, the other two must have a product of c.
Hmmm. Not what you wanted. So, divide by x+1, and you have
x^3 + ax^2 + bx + c
= (x+1)(x^2 + (a-1)x + (b-a+1))
So, the other two roots must have a product of b-a+1. Also, of course, you need c-b+a-1 = 0.
Check: Pick a,b,c such that a+c=b+1
For example, a=5,c=3,b=7
x^3+5x^2+7x+3 = (x+1)(x+1)(x+3)
and the other two roots multiply to 3, and 7-5+1 = 3.
Hmmm. Not what you wanted. So, divide by x+1, and you have
x^3 + ax^2 + bx + c
= (x+1)(x^2 + (a-1)x + (b-a+1))
So, the other two roots must have a product of b-a+1. Also, of course, you need c-b+a-1 = 0.
Check: Pick a,b,c such that a+c=b+1
For example, a=5,c=3,b=7
x^3+5x^2+7x+3 = (x+1)(x+1)(x+3)
and the other two roots multiply to 3, and 7-5+1 = 3.
Answered by
Simona
when u divide the cubic polynomial by x+1 u shud get remainder as 0 bcoz given that -1 is a zero..
When u divide the cubic polynomial ul get remainder as c-b+a-1 which is = 0
From here u can say that c=b-a+1
Therefore option numberA is correct
When u divide the cubic polynomial ul get remainder as c-b+a-1 which is = 0
From here u can say that c=b-a+1
Therefore option numberA is correct
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