Let p_1(x) = x^3 - 2020x^2 + b_1x + c_1 and p | vedclass

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(A) Let $p_1(x) = (x - \alpha)(x - \beta)(x - \gamma)$ and $p_2(x) = (x - \alpha)(x - \beta)(x - \delta)$.Given $p_1(x)q_1(x) + p_2(x)q_2(x) = x^2 - 3

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$p_1(3) + p_2(1) + 4028 = 0$

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(A) Let $p_1(x) = (x - \alpha)(x - \beta)(x - \gamma)$ and $p_2(x) = (x - \alpha)(x - \beta)(x - \delta)$.Given $p_1(x)q_1(x) + p_2(x)q_2(x) = x^2 - 3x + 2 = (x - 1)(x - 2)$.Since $p_1(x)$ and $p_2(x)$ share the factors $(x - \alpha)$ and $(x - \beta)$,the expression $(x - \alpha)(x - \beta)$ must divide $(x - 1)(x - 2)$.Thus,$\alpha = 1$ and $\beta = 2$.From the coefficients of $x^2$ in $p_1(x)$ and $p_2(x)$,we have $\alpha + \beta + \gamma = 2020 \implies 1 + 2 + \gamma = 2020 \implies \gamma = 2017$.Similarly,$\alpha + \beta + \delta = 2021 \implies 1 + 2 + \delta = 2021 \implies \delta = 2018$.So,$p_1(x) = (x - 1)(x - 2)(x - 2017)$ and $p_2(x) = (x - 1)(x - 2)(x - 2018)$.Calculating the values: $p_1(3) = (3 - 1)(3 - 2)(3 - 2017) = 2 	imes 1 	imes (-2014) = -4028$.$p_2(1) = (1 - 1)(1 - 2)(1 - 2018) = 0$.Therefore,$p_1(3) + p_2(1) + 4028 = -4028 + 0 + 4028 = 0$.

Let $p_1(x) = x^3 - 2020x^2 + b_1x + c_1$ and $p_2(x) = x^3 - 2021x^2 + b_2x + c_2$ be polynomials having two common roots $\alpha$ and $\beta$. Suppose there exist polynomials $q_1(x)$ and $q_2(x)$ such that $p_1(x)q_1(x) + p_2(x)q_2(x) = x^2 - 3x + 2$. Then the correct identity is

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