Let f ( x ) = x 3 + a x 2 + b x + c be polynomial with integer coefficients. If the roots of f ( x ) are integer and are in Arithmetic Progression, then ' a ' cannot take the value

Question

Quantrex Academy · Accepted Answer

Correct answer: B. 1214

Given that, f ( x ) = x 3 + a x 2 + b x + c Now, since the roots of the polynomial are in Arithmetic Progression, Let the roots are s - t , &#160; s , &#160; s + t &#8658; s - t + s + s + t = - a 1 &#8658; - 3 s = a Hence, a must be multiple of 3 . From given options we see that only 1214 is not multiple of 3 .

Let f ( x ) = x 3 + a x 2 + b x + c be polynomial with integer coefficients. If the roots of f ( x ) are integer and are in Arithmetic Progression, then ' a ' cannot take the value

Options

Correct answer

Step-by-step solution

More from Sequences and Series

Let f ( x ) = x 3 + a x 2 + b x + c be polynomial with integer coefficients. If the roots of f ( x ) are integer and are in Arithmetic Progression, then &#39; a &#39; cannot take the value

Options

Correct answer

Step-by-step solution

More from Sequences and Series

Let f ( x ) = x 3 + a x 2 + b x + c be polynomial with integer coefficients. If the roots of f ( x ) are integer and are in Arithmetic Progression, then ' a ' cannot take the value