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,

i) b-a+1
ii) b-a-1
iii) a-b+1
iv) a-b-1​

Answer 1

|| GIVEN ||

One of the zero of cubic polynomial = - 1

|| TO FIND ||

Product of the other two zeroes of Cubic polynomial.

|| SOLUTION ||

$p(x) = {x}^{3} + a {x}^{2} + bx + c$

Let us substitue the value of zero of cubic polynomial in the equation.

$\bf \: p(x) = {( - 1)}^{3} + a( { - 1)}^{2} + b( - 1) + c = 0$

$\implies - 1 + a - b + c = 0$

$\implies \: a - b + c = 0$

$\implies \: c = 1 - a + b$

Let α , β and γ be the zeroes of the cubic polynomial.

We Know that ,

$α \: \times β \times γ = \dfrac{ - d}{a}$

Since , d= c Let's Substitute a = 1

$\implies \: - 1 \times β \times γ = \dfrac{c}{1}$

→ βγ = c

From This C = 1 - a + b

c = b - a + 1 ( Option 1 )

For Further reference , See the attachment !

Answer 2

Answer:

Given :-

• If one of the zeros of the cubic polynomial x³ + ax² + bx + c is - 1.

To Find :-

• What is the product of the other two zeros.

Solution :-

Given equation :

$\longmapsto \sf\bold{\green{{x}^{3} + a{x}^{2} + bx + c}}$

Then,

$\implies \sf P(x) =\: {x}^{3} + a{x}^{2} + bx + c =\: 0$

By putting x = - 1 we get,

$\implies \sf P(x) =\: {(- 1}^{3} + a{(- 1)}^{2} + b(- 1)+ c =\: 0$

$\implies \sf (- 1)(- 1)(- 1) + a(- 1)(- 1) + b(- 1) + c =\: 0$

$\implies \sf (1)(- 1) + a(1) - b + c =\: 0$

$\implies \sf - 1 + a - b + c =\: 0$

$\implies \sf c =\: 1 - a + b$

$\implies \sf\bold{\purple{c =\: b - a + 1}}$

Now,

Let, α, β, γ be the roots of the cubic polynomial.

As we know that :

$\longmapsto \sf\boxed{\bold{\pink{\alpha + \beta + \gamma =\: \dfrac{- b}{a}}}}$

$\longmapsto \sf\boxed{\bold{\pink{\alpha\beta + \beta\gamma + \gamma\alpha =\: \dfrac{c}{a}}}}$

$\longmapsto \sf\boxed{\bold{\pink{\alpha\beta\gamma =\: \dfrac{- d}{a}}}}$

Since, we have to find the product,

Given :

$\mapsto \sf {x}^{3} + a{x}^{2} + bx + c =\: 0$

where,

• a = 1
• b = 1
• c = b
• d = c

So, by using the formula we get,

$\implies \sf \cancel{-} 1 \times \beta \times \gamma =\: \dfrac{\cancel{-} c}{1}$

$\implies \sf\bold{\blue{\beta\gamma =\: c}}$

Hence, we can write as :

$\implies\sf\bold{\red{c =\: b - a + 1}}$

$\therefore$ The product of the other two zeros is b - a + 1.

Hence, the correct options is option no (1) b - a + 1 .