2017/034) Let $p(x) = x^2 + bx +c $ where b and c are integers. if P(x) is a factor of both $x^4 + 6x^2 + 25$ and $3x^4+ 4x^2 + 28x + 5$, find the value of p(1)

Because P(x) devides $x^4 + 6x^2 + 25$ and  $3x^4+ 4x^2 + 28x + 5$ hence P(x) devides the GCD,
now $GCD(x^4 + 6x^2 + 25,3x^4+ 4x^2 + 28x + 5$
$= GCD(x^4 + 6x^2 + 25,3x^4+ 4x^2 + 28x + 5 - 3( x^4 + 6x^2 + 25))$
$= GCD(x^4 + 6x^2 + 25, -14x^2 + 28x -70))$
$= GCD(x^4 + 6x^2 + 25, x^2 -2x+5))$ deviding 2nd expression by -14
$= GCD(x^4 + 10x^2 + 25- 4x^2, x^2-2x + 5)$
$= GCD((x^2 + 5)^2- (2x)^2, x^2-2x + 5)$
$=GCD((x^2+2x+5)(x^2-2x+5),x^2-2x  + 5) $
$=x^2 - 2x + 5$
so as p(x) is of the form $x^2+bx+ c$ so $p(x) = x^2 - 2x+5$ and hence $p(1) = 4$