Question

If one of the zeros of the cubic polynomial
${x}^{3} + a {x}^{2} + bx + c$
is -1,then the product of the other two zeros is
a.)b-a+1
b.)b-a-1
c.)a-b+1
d.)a-b-1

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Answer 1

Answer:

For a cubic polynomial, we know that:

$\tt{\alpha + \beta + \gamma = \frac{-b}{a}}$ ..(1)

$\tt{\alpha \beta + \alpha \gamma + \beta \gamma = \frac{c}{a}}$ ..(2)

When $\tt{\gamma = -1}$

Using (1):

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

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

Using (2):

$\tt{\alpha \beta - \alpha - \beta = b}$

=> $\tt{\alpha \beta = b + \alpha + \beta}$

Put (i):

=> $\tt{\alpha \beta = b + 1 - a}$

Answer: (a) b - a + 1

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