If one of the zeros of the cubic polynomial

Question:

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

(a) a − b − 1
(b) − a − 1
(c) 1 − a + b
(d) 1 + a − b

Solution:

(c) $1-a+b$

Since $-1$ is a zero of $x^{3}+a x^{2}+b x+c$, we have:

$(-1)^{3}+a \times(-1)^{2}+b \times(-1)+c=0$

$=>a-b+c-1=0$

$=>c=1-a+b$

Also, product of all zeroes is given by

$\alpha \beta \times(-1)=-c$

$=>\alpha \beta=c$

$=>\alpha \beta=1-a+b$

 

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