Question

(8r^2-12r+4)-(3r^2+5r-1) in standard form

Answer 1

Answer:

$=5r^2-17r+5$

Step-by-step explanation:

$\mathrm{Apply\:rule}:\quad \left(a\right)=a$

$\left(8r^2-12r+4\right)=8r^2-12r+4$

$=8r^2-12r+4-\left(3r^2+5r-1\right)$

$\mathrm{Apply\:the\:distributive\:law}:\quad \:-\left(a+b\right)=-a-b$

$-\left(3r^2+5r-1\right)=-3r^2-5r+1$

$=8r^2-12r+4-3r^2-5r+1$

$\mathrm{Group\:like\:terms}$

$=8r^2-3r^2-12r-5r+4+1$

$\mathrm{Add\:the\:numbers:}\:4+1=5$

$=8r^2-3r^2-12r-5r+5$

$\mathrm{Add\:similar\:elements:}\:8r^2-3r^2=5r^2$

$=5r^2-12r-5r+5$

$\mathrm{Add\:similar\:elements:}\:-12r-5r=-17r$

$=5r^2-17r+5$