Question

If one of the zeroes of the cubic polynomial x3 + ax2 + bx + c is –1, then the Product of the other two zeroes is

Answer 1

Question:

If one of the zeroes of the cubic polynomial x³ + ax² + bx + c is –1, then the Product of the other two zeroes is

Answer:

The Product of the other two zeroes is

➞ b - a + 1

Given:

▪ The cubic polynomial is : x³ + ax² + bx + c

▪ One of the zeroes of the cubic polynomial is : -1

To Find:

Product of other two zeroes of the polynomial

Solution:

We are given,

▪ The cubic polynomial is : x³ + ax² + bx + c

▪ One of the zeroes of the cubic polynomial is : -1

Let α , β be the other zeros of the given

polynomial x³ + ax² + bx + c.

$\rm Sum\:of\:the\:zeroes\: = \dfrac{ - coefficient \: of \: {x}^{2}} {coefficient\:of \:{x}^{3}}$

${-1 + \alpha + \beta = - \dfrac{a}{1} = - a}$

${\alpha + \beta = - a + 1 ..................(1)}$

Again,

$\rm ( - 1) \alpha + \alpha \beta + ( - 1) \beta=\dfrac{-coefficient \: of \: x} {coefficient\:of\:{x}^{3}}$

${\rm{ - \alpha + \alpha \beta - \beta = \dfrac {-b}{a} }}$

${\rm{ \alpha \beta = b + \alpha + \beta} }$

${\rm{( \alpha + \beta = - a + 1\: from \ equation \: (1))}}$

$b - a + 1$

⛬ The Product of the other two zeroes is

➞ b - a + 1

Answer 2

Answer:

b-a+1

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,

product of all zeroes = constant term/coefficient of x³ =c/1

αβγ = -c

⇒ (-1)βγ = −c [∴α = -1]

⇒ βγ = c

⇒ βγ = 1 -a + b

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,

⇒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