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If one of the zeroes of the cubic polynomial $x^{3}+a x^{2}+b x+c$ is $-1$, then the product of the other two zeroes is (a) $a-b-1$ (b) $b-a-1$ (c) $1-a+b$ (d) $1+a-b$
If one of the zeroes of the cubic polynomial $x^{3}+a x^{2}+b x+c$ is $-1$, then the product of the other two zeroes is (a) $a-b-1$ (b) $b-a-1$ (c) $1-a+b$ (d) $1+a-b$
CBSE | Class 10 | Excercise 2D | Maths | Polynomials | RS Aggarwal
If one of the zeroes of the cubic polynomial $x^{3}+a x^{2}+b x+c$ is $-1$, then the product of the other two zeroes is (a) $a-b-1$ (b) $b-a-1$ (c) $1-a+b$ (d) $1+a-b$

The correct option is option (c) $1-\mathrm{a}+\mathrm{b}$

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

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

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

$\Rightarrow \mathrm{c}=1-\mathrm{a}+\mathrm{b}$

product of all zeroes is given by

$\alpha \beta \times(-1)=-\mathrm{c}$

$\Rightarrow \alpha \beta=\mathrm{c}$

$\Rightarrow \alpha \beta=1-\mathrm{a}+\mathrm{b}$

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