# Show that if $p(x)$ is not divisible over integral domain then $p(x+a)$ is also not divisible. [duplicate]

Let $$a\in R$$. How to show that if a polynomial $$p(x)\in R[X]$$ where $$R$$ is an integral domain is not divisible if and only if $$p(x+a)$$ is not divisible?

## marked as duplicate by Bill Dubuque divisibility StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Feb 12 at 17:16

Hint: $$p(x) = q(x)r(x)$$ if, and only if, $$p(x + a) = q(x+a)r(x+a)$$ (that is, $$p(x)$$ is reducible if and only if $$p(x+a)$$ is reducible). Can you fill the gaps?