# General Multiplication of Polynomials

Alas, there are no quick and easy patterns to use when we're multiplying any two polynomials that don't fit the description of one of our special cases. We just need to apply the distributive property and apply the distributive property again. And maybe once more, being careful to write down all the intermediate work. As the polynomials become more complex, there's more work to write down, but the problems don't become *harder*...just *longer*. You may want to free up some time in your afternoon schedule.

### Sample Problem

Find (5*x*^{2} + 3*x* + 7)(-2*x*^{3} – 4*x*^{2} + 3*x* + 4).

Hang on, here we go. First application of the distributive property:

(5*x*^{2} + 3*x* + 7)(-2*x*^{3}) + (5*x*^{2} + 3*x* + 7)(-4*x*^{2}) + (5*x*^{2} + 3*x* + 7)(3*x*) + (5*x*^{2} + 3*x* + 7)(4)

In English you can have a run-on sentence; we're pretty sure this is a run-on expression. It's too monstrous for one line, so we'll break it into several:

(5*x*^{2} + 3*x* + 7)(-2*x*^{3})

+ (5*x*^{2} + 3*x* + 7)(-4*x*^{2})

+ (5*x*^{2} + 3*x* + 7)(3*x*)

+ (5*x*^{2} + 3*x* + 7)(4)

Now apply the distributive property to each line and simplify. Quickly, before this expression thinks of something else it wants to say.

The first line will go like this:

(5x^{2} + 3x + 7)(-2x^{3}) | = | (5x^{2})(-2x^{3}) + (3x)(-2x^{3}) + 7(-2x^{3}) |

= | -10x^{5} – 6x^{4} – 14x^{3} |

After we're done following the same steps for each of the other lines, we'll find this sum:

-10x^{5} | – 6x^{4} | – 14x^{3} | |||

– 20x^{4} | – 12x^{3} | – 28x^{2} | |||

+ 15x^{3} | + 9x^{2} | + 21x | |||

+ 20x^{2} | + 12x | + 28 |

To make things as un-monstrous as possible, we've lined everything up so like terms are stacked on top of each other. It's like Bejeweled, but with coefficients, variables, and exponents!

To find our final answer, we add down the columns:

-10*x*^{5} – 26*x*^{4} – 11*x*^{3} + *x*^{2} + 33*x* + 28

Hopefully you won't be asked to do too many problems like this, but if you are, you'll know what to do.

Another thing we can do that's more messy than it is difficult is multiplying three or more polynomials together. We promise no run-on expressions this time.

### Sample Problem

Find (*x* + 1)(2*x* + 3)(*x*^{2} + *x*).

To do this problem, break it into pieces so that we only need to multiply two polynomials together at a time. Any time you can break something down into simpler parts, we encourage you to do so. Unless it's your car. That should probably stay in one piece.

First we use the distributive property to find the product of the first two polynomials:

(*x* + 1)(2*x* + 3) = 2*x*^{2} + 5*x* + 3

Then we find (2*x*^{2} + 5*x* + 3)(*x*^{2} +* x*).

Apply the distributive property once to get:

(2*x*^{2} + 5*x* + 3)(*x*^{2})

+ (2*x*^{2} + 5*x* + 3)(*x*)

Then apply the distributive property to each line and simplify:

2x^{4} | + 5x^{3} | + 3x^{2} | |

+ 2x^{3} | + 5x^{2} | + 3x |

Add down the columns to find the final answer:

2x^{4} | + 7x^{3} | + 8x^{2} | + 3x |

Whenever we're multiplying together more than three polynomials, it doesn't matter which ones we multiply first. We could choose the last two, or the first and last, or whichever two we want. This may be one of the only times in algebra that you're able to exercise some degree of freedom or creativity, so live it up.

However, there's often a rhyme and reason to it. Sometimes we choose which two polynomials to multiply first because we know their product will be nice and attractive. Basically, we choose our polynomials like we choose our prom dates.