Question 404630
(x-1) (x+3) (x-3)
It will require two stages to multiply this out:<ol><li>Multiply two of these three factors. (And since multiplication is Commutative and Associative we can pick any two factors to multiply to start!)</li><li>Multiply the answer from step 1 by the remaining factor.</li></ol>
Since we get to pick which pair of factors to multiply I am going to pick the easiest pair. The last two factors, (x+3) and (x-3), fit the pattern for {{{(a+b)(a-b) = a^2-b^2}}}. So multiplying them will be easy. Using the pattern, with "a" being x and "b" being 3 we get:
{{{(x-1) ((x)^2 - (3)^2)}}}
or
{{{(x-1) (x^2 - 9)}}}
Now we multiply what is left. There is no pattern for this. Since it is one binmial times another we can use FOIL. But before that I like to rewrite the subtractions as additions. Subtractions are a common source of confusion and errors.
{{{(x+ (-1)) (x^2 + (-9))}}}
Now we'll use FOIL:
{{{x*x^2 + x*(-9) + (-1)*x^2 + (-1)(-9)}}}
which simplifies as follows:
{{{x^3 + (-9x) + (-x^2) + 9}}}
Usually terms are put in descending order of the exponents so this would become:
{{{x^3 + (-x^2) + (-9x) + 9}}}