How do you multiply #(x - 1)(x + 1)(x - 3)(x + 3)#?
1 Answer
May 8, 2016
Explanation:
Note the difference of squares identity:
#a^2-b^2=(a-b)(a+b)#
So we find:
#(x-1)(x+1) = x^2-1^2 = x^2-1#
#(x-3)(x+3) = x^2-3^2 = x^2-9#
Then if it helps, we can use FOIL to multiply out the resulting two binomials:
#(x^2-1)(x^2-9)#
#=overbrace((x^2 * x^2))^"First" + overbrace(((x^2) * (-9)))^"Outside" + overbrace(((-1) * (x^2)))^"Inside" + overbrace(((-1) * (-9)))^"Last"#
#=x^4-9x^2-x^2+9#
#=x^4-(9+1)x^2+9#
#=x^4-10x^2+9#