## Precalculus: Mathematics for Calculus, 7th Edition

quadratic x+1 $W^{2}-5W+6=0$
$(x+1)^2-5(x+1)+6=0$ Remember Linear takes form of $ax+b=0$ quadratic takes form of $ax^{2}+bx+c=0$ so this is quadratic type Set W=(x+1) [substitute (x+1) with W] $W^2-5W+6=0$