p(x) is a polynomial such that p(x)= (x-a)(x+10) + 1, a is not equal to 0. 'a' attains integral values for integral roots of the above polynomial. what is the sum of all posible valueesof 'a'??

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$Consider the following polynomial . p (x) = (x - a) (x + 10) + 1, a \neq 0 a \in Z (non zero Integers) Write the above polynomial in general form . p (x) = (x - a) (x + 10) + 1 = x^{2} + 10 x - ax - 10 a + 1 = x^{2} + (10 - a) x + (1 - 10 a) For a = 1, 2, 3, 4, 5, . . . . the corresponding polynomials are p (x) = x^{2} + 9 x - 9, a = 1 p (x) = x^{2} + 8 x - 19, a = 2 p (x) = x^{2} + 7 x - 29, a = 3 p (x) = x^{2} + 6 x - 39, a = 4 p (x) = x^{2} + 5 x - 49, a = 5 ⋮ If α and β are roots of the above equations . Then it implies that, α + β = 9, α β = - 9, a = 1 α + β = 8, α β = - 19, a = 2 α + β = 7, α β = - 29, a = 3 α + β = 6, α β = - 39, a = 4 α + β = 5, α β = - 49, a = 5 Note that no integral values of α and β satisfy the above relations . Hence the roots are not integral values for a = 1, 2, 3, 4, 5, . . . . For a = - 1, - 2, - 3, - 4, - 5, . . . . the corresponding polynomials are p (x) = x^{2} + 11 x + 11, a = 1 p (x) = x^{2} + 12 x + 21, a = 2 p (x) = x^{2} + 13 x + 31, a = 3 p (x) = x^{2} + 14 x + 41, a = 4 p (x) = x^{2} + 15 x + 51, a = 5 ⋮ If α and β are roots of the above equations . Then it implies that, α + β = 11, α β = 11, a = 1 α + β = 12, α β = 21, a = 2 α + β = 13, α β = 31, a = 3 α + β = 14, α β = 41, a = 4 α + β = 15, α β = 51, a = 5 Again, note that no integral values of α and β satisfy the above relations . Such α and β does not exist . Hence the roots are not integral values for a = - 1, - 2, - 3, - 4, - 5, . . . . Therefore, there does not exist any integral value of " a " for which the given equation has integral roots$

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