## Precalculus: Mathematics for Calculus, 7th Edition

$(A + B)^{2}$ = $A^2$ + 2AB + $B^2$ $(2x + 3)^2$ = 4$x^2$ + 6x + 9
When you square a term, you just multiply it by itself, therefore: $(A + B)^{2}$ = (A + B)(A + B) Multiplying these by FOIL (first, outer, inner, last) yields $A^2$ + AB + AB + $B^2$, which is equvalent to $A^2$ + 2AB + $B^2$ For part two, A = 2x and B = 3, therefore this squared is: $A^2$ + 2AB + $B^2$ $(2x)^2$ + 2(2x)(3) + $3^2$ 4$x^2$ + 6x + 9