How do you find the product of #(x-3)^2#?

2 Answers
Apr 4, 2018

#(x-3)^2=(x-3)(x-3)#

From here we can just FOIL; first, outer, inner, last. Multiply the first term in the first bracket (x), by the FIRST term in the second bracket (x), giving us #x^2#. Then, we multiply the outermost term on the left side by the OUTERmost on the right side, giving us #-3 * x#, equal to #-3x#. We then multiply the INNERmost terms together, which are again -3 and #x#, giving us #-3x# again. Finally we multiply the LAST term in the first bracket by the last term of the second, giving us #-3 * -3#, or 9. Then, we simply write these out, in order:

#x^2-3x-3x+9#

Collect like terms:

#x^2-6x+9#

And you're done!

Apr 4, 2018

Answer:

#x^3-6x+9#

Explanation:

  1. Let's say #A=x-3# (this is to help you understand).
    So then we would have #A^2#, which gets you #AxxA#.

  2. Now let's apply that back into the original problem.
    #(x-3)^2=(x-3)xx(x-3)=(x-3)(x-3)#

  3. We can use the Distributive Property to make this more easily readable:
    #\color(red)(x(x-3))+\color(green)((-3)(x-3))#
    Simplify by re-applying Distributive Property:
    #\color(red)(x(x)+x(-3))+\color(green)((-3)(x)+(-3)(-3))#

  4. And multiply...
    #x^2+(-3x)+(-3x)+9=x^3-6x+9#

#\color(tomato)(\text(For the third step:))# This is really just me explaining FOIL in a different way.