# Multiplication

Multiplying polynomials is just an application of the distributive property. The whole secret lies in being *very* careful to make sure you don't mix up any of the terms since the process can get pretty messy. The process goes like this:

Multiply: | $(3x-4)(x+2)$ |

Take the whole first polynomial and distribute it into the second one. | $(3x - 4)(x) + (3x - 4)(2)$ |

It's a little unusual to put the smaller term at the end so I'm going to rearrange the previous expression so it looks a little more natural. | $x(3x - 4) + 2(3x - 4)$ |

Now we're going to apply the Distributive Property again. First, I'm going to distribute the x into the (3x - 4). | $3x \cdot x - 4x + 2(3x - 4)$ |

Now I'll do the same thing with the 2 and the (3x - 4). | $3x \cdot x - 4x + 2 \cdot 3x - 2 \cdot 4$ |

Next I'm going to multiply the x's and the numbers together. | $3x^2 - 4x + 6x - 8$ |

And, finally, I'll combine the like terms. | $3x^2 + 2x - 8$ |

That's the formal way of doing multipication that will work with literally any two polynomials. If you've taken a basic algebra class, you may have heard of another approach that's referred to as the FOIL method. To be perfectly honest, I'm not a fan of the process and would encourage you not to bother with it. Ultimately, it's just another procedure for you to learn that only works in the special case where you're multiplying two binomials. My opinion is that you're better off learning the method I described above since it's no more complicated and works in every possible case.

Now that I've gotten that off my chest, let's look at some more examples

## Quick Tip - FOIL

Okay - I know I said I've got no use for 'FOIL' but I'm sure you've seen it in class or in a textbook so I feel obligated to discuss it.

FOIL is an acronomy for First, Outer, Inner, Last. What it means is that you can multiply two binomials together by multipying their first terms, their 'inner' terms, their 'outer terms' and their last terms and then adding the results together. In practice it goes like this: If I asked you to multiply:

$(2x+1)(x^2-2)$

You would start by calculating these parts:

first terms | $2x \cdot x^2 = 2x^3$ |

'outer' terms | $2x \cdot (-2) = -4x$ |

'inner' terms | $1 \cdot x^2 = x^2$ |

last terms | $1 \cdot (-2) = -2$ |

Our final answer is the sum of those four terms:

$(2x+1)(x^2-2)=2x^3+x^2-4x-2$

# Example 1

**Simplify $(-x + 2)(-3x - 2)$.**

This question is a little trick since there are a lot of negative signs floating around. Pay special attention to how I keep track of them as we go along.

First, I'm going to distribute the $(-x + 2)$ into the second set of parentheses.

$-3x(-x + 2) - 2(-x + 2)$

I went ahead and kept the terms of the second parentheses in front since that's a more natural way of writing the expressions and helps to keep the negative signs in the right place. Now I'm going to distribute the -3x and the -2 into the parentheses that follow them:

$-3x(-x) + 2(-3x) - 2(-x) - 2(2)$

Notice how I kept the negative signs with the terms as I distributed them. Now I'll simplify the individual terms by combining the x's and multiplying out the numbers.

$3x^2 - 6x + 2x - 4$

Be sure you kept track of what happened to the negative signs. A number of them became positive since we had two negatives being multiplied together. Finally, I'll combine the like terms to get our final answer.

$3x^2 - 4x - 4$

# Example 2

**Simplify $(-x^2 - x + 2)(-3x^4 + 3x^3 - 2)$.**

This one is complicated only by the number of terms in each polynomial. We'll still handle it exactly the same way we did the previous examples: Take every term in the first polynomial and multiply them by every term in the second. For bigger problems like this, I like to use a column approach like we did with adding.

$-x^2(-3x^4 + 3x^3 - 2)$ | $ = $ | $(-x^2)(-3x^4) + (-x^2)(3x^3) + (-2)(-x^2)$ | $ = $ | $3x^6$ | $ - 3x^5$ | $ + 2x^2$ | ||||

$-x(-3x^4 + 3x^3 - 2)$ | $ = $ | $(-x)(-3x^4) + (-x)(3x^3) + (-2)(-x)$ | $ = $ | $3x^5$ | $ - 3x^4$ | $ + 2x$ | ||||

$2(-3x^4 + 3x^3 - 2)$ | $ = $ | $2(-3x^4) + 2(3x^3) - 2(2)$ | $ = $ | $-6x^4$ | $ + 6x^3$ | $ - 4$ | ||||

Now, to get our final answer, all we have to do is add down the columns: |
$3x^6$ | $ + (-3 + 3)x^5$ | $ + (-3 - 6)x^4$ | $ + 5x^3$ | $ + 2x^2$ | $ + 2x$ | $ - 4$ | |||

$3x^6 - 9x^4 + 5x^3 + 2x^2 + 2x - 4$ |