## Algebra 1: Common Core (15th Edition)

Error: $(9x)^{2}\neq 9x^{2}.$ The correct term is $3x,$ because $(3x)^{2}=9x^{2}.$ Correct factorization: $(3x+7)(3x-7)$
The square of 7 is 49, but the square of $9x$ is $81x^{2}$, not $9x^{2}.\\\\$ In applying the formula $a^{2}-b^{2}=(a+b)(a-b)$, here, $a=3x$, and $b=7$ So, the correct way to factor is $9x^{2}-49=(3x+7)(3x-7)$