Faktorisering av andregradsuttrykk ved å bruke kvadratsetningene

$=\left(x+1\right)\left(x-1\right)$

$=\left(x+2\right)\left(x-2\right)$

$=\left(x+3\right)\left(x-3\right)$

$=\left(x+4\right)\left(x-4\right)$

$=\left(x+5\right)\left(x-5\right)$

$=\left(x+6\right)\left(x-6\right)$

$=\left(x+7\right)\left(x-7\right)$

$=\left(x+8\right)\left(x-8\right)$

$=\left(x+9\right)\left(x-9\right)$

$=\left(x+10\right)\left(x-10\right)$

$=\left(x+11\right)\left(x-11\right)$

$=\left(x+12\right)\left(x-12\right)$

$=\left(2x+5\right)\left(2x-5\right)$

$=2\left(x^2-9\right)=2\left(x+3\right)\left(x-3\right)$

$=3x\left(x-16\right)$

$=2\left(9-x^2\right)=2\left(3+x\right)\left(3-x\right)$

$=\left(x+\sqrt{5}\right)\left(x-\sqrt{5}\right)$

$=7\left(x^2-3\right)=7\left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right)$

$=x^2-2·x·1+1^2={\left(x-1\right)}^2$

$=x^2-2·x·7+7^2={\left(x-7\right)}^2$

$=x^2-2·x·3+3^2={\left(x-3\right)}^2$

$=4\left(9+6b+b^2\right)=4\left(3^2+2·3·b+b^2\right)=4{\left(3+b\right)}^2$

$={\left(x-2\right)}^2-6^2=\left(\left(x-2\right)+6\right)·\left(\left(x-2\right)-6\right)=\left(x+4\right)\left(x-8\right)$