If one of the zeroes of the cubic polynomial `x^3 + ax^2 + bx + c` is `-1`,then find the product of other two zeroes.

Options

A. `b - a+1`
B. `b-a-1`
C. `a-b+1`
D. `a-b-1`

Answer

A

Explanation

Let `p(x) =x^(3) +ax^(2) +bx +c`
Let `alpha, beta` and `gamma` be the zeroes of the given cubic polynomial `p(x)`.
`:. alpha =- 1`" "[given]
and `p(-1) = 0`
`rArr (-1)^(3) +a(-1)^(2)+b(-1)+c = 0`
`rArr -1 +a -b +c = 0`
`rArr c = 1 - a+b` ...(i)
We know that,
Product of all zeroes `= (-1)^(3) ("Constant term")/("Coefficient of" x^(3)) =- (c)/(1)`
`alpha beta gamma =- c`
`rarr (-1) beta gamma =- c " "[ :' alpha =- 1]`
`rArr beta gama = c`
`rArr beta gamma = 1 - a+b` " "[from Eq.(i)]
Hence, product the other two roots is `1-a +b`.
Alternate Method
Since, -1 is one of the zeroes of the cubic polynomal `f(x) = x^(2)+ax^(2)+bx +c` i.e., `(x+1)` is a factor of `f(x)`.
Now, using division algorithm,
`{:(" "ul(x^(2)+(a-1)x+(b-a+1))),( {:x+1) " "x^(3)+ax^(2)+bx+c),(" "ul(x^(3)+x^(2)" ")),(" "(a-1)x^(2)+bx),(" "ul((a-1)x^(2)+(a-1)x)),(" "(b-a+1)x+c),(" "ul((b-a+1)x(b-a+1))),(" "(c-b+a-1)):}`
`:. x^(3) +ax^(2) +bx +c =(x+1) xx {x^(3)+(a-1)x+(b-a+1)} +(c-b+a-1)`
`rArr x^(3)=ax^(2)+bx +(b-a+1) =(x+1) {x^(2)+(a-1)x+(x+(b-a+1)}`
Let `alpha` and `beta` be the other two zeroess of the given polnomial, then
Product of zeroes `= (-1) alpha. beta = (-"Constant term")/("Coefficient of" x^(3))`
`rArr - alpha. beta =(-b-a+1)/(1)`
`rArr alpha beta = - a +b +1`
Hence, the required product of other two roots is `(-a+b+1)`.