Q. 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) Let p(x) = x3 + ax2 + bx + c

Let a, p and y be the zeroes of the given cubic polynomial p(x).

∴  α = -1                                        [given]

and p(−1) = 0

⇒ (-1)3 + a(-1)2 + b(-1) + c = 0

⇒ -1 + a- b + c = 0

⇒ c = 1 -a + b                                                            …(i)

We know that,

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refer to attatchment#1

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αβγ = -c

⇒ (-1)βγ = −c                                                                             [∴α = -1]

⇒ βγ = c

⇒ βγ = 1 -a + b                                                                [from Eq. (i)]

Hence, product of the other two roots is 1 -a + b.

Alternate Method

Since, -1 is one of the zeroes of the cubic polynomial f(x) = x2 + ax2 + bx + c i.e., (x + 1) is a factor of f{x).

Now, using division algorithm,

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refer to attatchment #2

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⇒x3 + ax2 + bx +c = (x + 1) x {x2 + (a – 1)x + (b – a + 1)> + (c – b + a -1)

⇒x3 + ax2 + bx + (b – a + 1) = (x + 1) {x2 + (a – 1)x + (b -a+ 1)}  

Let a and p be the other two zeroes of the given polynomial, then

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refer to attatchment #3

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