Question

If one of the zeros of the cubic polynomial x3 + ax2 + bx + C is - 1, then the product of other two zeros is

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Answer 1

(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,

$\sf{Product\;of\;zeroes = (-1)^{3}\times \frac{constant\;term}{coefficient\;of\;x^{3}}=-\frac{c}{1}}$

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.

Answer 2

Step-by-step explanation:

(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,

\sf{Product\;of\;zeroes = (-1)^{3}\times \frac{constant\;term}{coefficient\;of\;x^{3}}=-\frac{c}{1}}Productofzeroes=(−1)

3

×

coefficientofx

3

constantterm

=−

1

c

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.