Odd Prime Cubic Root

Algebra Level pending

Consider the general cubic polynomial with nonzero integer coefficients and constant term, namely, f ( x ) = a x 3 + b x 2 + c x + d f(x) = ax^3 +bx^2 +cx +d where a > 2 a>2 is prime, d > 2 |d|>2 is prime, and f ( 1 ) = a + d 0 f(1) = a+d\neq0 . From the given info above, c = b c = -b . Find another expression of c c that proves f ( x ) f(x) has root x = 1 x=-1 .

c = d f ( 1 ) 2 c=d-\frac{f(1)}{2} c = a + f ( 1 ) 2 c=a+\frac{f(1)}{2} c = a f ( 1 ) 2 c=a-\frac{f(1)}{2} c = d + f ( 1 ) 2 c=d+\frac{f(1)}{2}

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1 solution

Frank Giordano
Mar 2, 2017

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