## College Algebra 7th Edition

$x=\pm1$
$x^{8}+15x^{4}=16$ $x^{8}+15x^{4}-16=0$ $(x^{4}+16)(x^{4}-1)=0$ $x^{4}+16=0$ or $x^{4}-1=0$ $x^4=-16$ or $x^4=1$ $x=\pm\sqrt[4]{-16}$ or $x=\pm\sqrt[4]{1}=\pm1$ The left solution is not real, so we ignore it. Thus the solution is $x=\pm1$