If one of the zeroes of the cubic polynomial | vedclass

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(D) Let the cubic polynomial be $p(x) = x^{3} + a x^{2} + b x + c$.Let the zeroes of the polynomial be $\alpha, \beta,$ and $\gamma$.Given that one ze

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$b-a+1$

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(D) Let the cubic polynomial be $p(x) = x^{3} + a x^{2} + b x + c$.Let the zeroes of the polynomial be $\alpha, \beta,$ and $\gamma$.Given that one zero is $\alpha = -1$.Since $\alpha = -1$ is a zero,$p(-1) = 0$.Substituting $x = -1$ into the polynomial:$(-1)^{3} + a(-1)^{2} + b(-1) + c = 0$$-1 + a - b + c = 0$$c = 1 - a + b$ ... $(i)$We know that for a cubic polynomial $Ax^{3} + Bx^{2} + Cx + D$,the product of the zeroes is given by $-\frac{D}{A}$.Here,the product of the zeroes is $\alpha \cdot \beta \cdot \gamma = -\frac{c}{1} = -c$.Substituting $\alpha = -1$:$(-1) \cdot \beta \cdot \gamma = -c$$\beta \cdot \gamma = c$.Substituting the value of $c$ from equation $(i)$:$\beta \cdot \gamma = 1 - a + b$.Thus,the product of the other two zeroes is $b - a + 1$.

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

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