Distribute the referral code to your friends and ask them to register with Tutorix using this referral code. Once we get 15 subscriptions with your referral code, we will activate your 1 year subscription absolutely free. Your subscribed friend will also get 1 month subscription absolutely free. Open in App Suggest Corrections 1 Q. By divisible algorithm, when  Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, (i) x3 − 2ax2 + ax − 1 (ii) x5 − 3x4 − ax3 + 3ax2 + 2ax + 4 In each of the following two polynomials, find the value of a, if x − a is factor:(i) x6 − ax5 + x4 − ax3 + 3x − a + 2 (ii x5 − a2x3 + 2x + a + 1) Find the values of a and b so that (x + 1) and (x − 1) are factors of x4 + ax3 − 3x2 + 2x + b.Find the value k if x − 3 is a factor of k2x3 − kx2 + 3kx − k.f(x) = x3 − 6x2 + 11x − 6, g(x) = x2 − 3x + 2Q. By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, (i) x3 − 2ax2 + ax − 1 (ii) x5 − 3x4 − ax3 + 3ax2 + 2ax + 4 In each of the following two polynomials, find the value of a, if x − a is factor:(i) x6 − ax5 + x4 − ax3 + 3x − a + 2 (ii x5 − a2x3 + 2x + a + 1) Find the values of a and b so that (x + 1) and (x − 1) are factors of x4 + ax3 − 3x2 + 2x + b.Find the value k if x − 3 is a factor of k2x3 − kx2 + 3kx − k.f(x) = x3 − 6x2 + 11x − 6, g(x) = x2 − 3x + 2Q. By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, (i) x3 − 2ax2 + ax − 1 (ii) x5 − 3x4 − ax3 + 3ax2 + 2ax + 4 In each of the following two polynomials, find the value of a, if x − a is factor:(i) x6 − ax5 + x4 − ax3 + 3x − a + 2 (ii x5 − a2x3 + 2x + a + 1) Find the values of a and b so that (x + 1) and (x − 1) are factors of x4 + ax3 − 3x2 + 2x + b.Find the value k if x − 3 is a factor of k2x3 − kx2 + 3kx − k.f(x) = x3 − 6x2 + 11x − 6, g(x) = x2 − 3x + 2Q. By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, By divisible algorithm, when Let Let We have, Clearly, Therefore, and Adding (i) and (ii), we get, Putting this value in equation (i), we get, Hence, (i) x3 − 2ax2 + ax − 1 (ii) x5 − 3x4 − ax3 + 3ax2 + 2ax + 4 In each of the following two polynomials, find the value of a, if x − a is factor:(i) x6 − ax5 + x4 − ax3 + 3x − a + 2 (ii x5 − a2x3 + 2x + a + 1) Find the values of a and b so that (x + 1) and (x − 1) are factors of x4 + ax3 − 3x2 + 2x + b.Find the value k if x − 3 is a factor of k2x3 − kx2 + 3kx − k.f(x) = x3 − 6x2 + 11x − 6, g(x) = x2 − 3x + 2Q. 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What must be subtracted from X 3 6x 2 15x 80 so that the result is exactly divisible by X 2 X 12 degrees?
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