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![\bf p(x)=x^4-256\qquad \boxed{2^8=256}\qquad p(x)=x^{2\cdot 2}-2^{4\cdot 2} \\\\\\ p(x)=(x^2)^2-(2^4)^2\implies p(x)=[x^2-2^4]~[x^2+2^4] \\\\\\ p(x)=[x^2-(2^2)^2]~[x^2+2^4]\implies p(x)=[(x-2^2)(x+2^2)]~[x^2+2^4] \\\\\\ p(x)=(x-4)(x+4)(x^2+16)](/tpl/images/0143/8399/51383.png)
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Factor this completely p(x)=x^4-256...
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