Middle School Math Solutions – Expand Calculator, Polynomial Multiplication

As of right now, you should be familiar with the distributive law and the FOIL method. We will now use both of them and apply them to our next topic, polynomial multiplication.

Recall: A polynomial is an expression of many (poly-) terms (-nomial) that are added and/or subtracted together. The terms include coefficients, variables, and POSITIVE exponents.

Examples: 4x^2 ,2x^3+5x^5 ,\frac{2}{3}x-1

There are a variety of polynomial multiplication problems. They might look intimidating at first, but we will go over how to solve them step by step using the distributive law and FOIL method, so you can see how to do them easily.

First example (click here):

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

1. Distribute the 4x

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

We applied the distributive law here for a polynomial with three terms                  (a(b+c+d)=ab+ac+ad)

2. Simplify

                                                4x(3x^2+2x-1)=12x^3+8x^2-4x

We multiplied the terms.

That one wasn’t bad. Let’s see another one.

Next example (click here):

                                                            (2x+1)(4x+3y+2)

1. Distribute 2x+1

                                      (2x+1)(4x+3y+2)=2x(4x+3y+2)+1(4x+3y+2)

We distributed 2x+1 to make the problem easier for us to solve.

2. Use the distributive law

                                              2x(4x+3y+2)=2x∙4x+2x∙3y+2x∙2

                                                  1(4x+3y+2)=4x∙1+3y∙1+2∙1

We plug these values back in

                              (2x+1)(4x+3y+2)=2x∙4x+2x∙3y+2x∙2+4x∙1+3y∙1+2∙1

3. Simplify

                                         (2x+1)(4x+3y+2)=8x^2+6xy+4x+4x+3y+2
                                                           =8x^2+6xy+8x+3y+2

Last example (click here):

                                                                (x+1)(x+2)(x+3)

1. Apply the FOIL method to two of the polynomials

                                                      (x+1)(x+2)=x∙x+x∙2+1∙x+1∙2
                                                                     =x^2+3x+2

2. Plug x^2+3x+2 in for (x+1)(x+2)

                                                  (x+1)(x+2)(x+3)=(x^2+3x+2)(x+3)

Now this problem looks like the prior example.

3. Distribute x+3

                                           (x+1)(x+2)(x+3)=x(x^2+3x+2)+3(x^2+3x+2)

I always prefer to distribute the polynomial with the smallest terms, but you can pick to distribute the quadratic. In that case you would have:

                                           (x^2+3x+2)(x+3)=x^2 (x+3)+3x(x+3)+2(x+3)

4. Use the distributive law

                                            (x+1)(x+2)(x+3)=x(x^2+3x+2)+3(x^2+3x+2)
                                                   =x∙x^2+x∙3x+x∙2+3∙x^2+3∙3x+3∙2

5. Simplify

                                             (x+1)(x+2)(x+3)=x^3+3x^2+2x+3x^2+9x+6
                                                                    =x^3+6x^2+11x+6

Not bad! You can see when we break the problem down, step by step, it is a lot easier to solve because we use the basics of expanding (distributive law and FOIL). For more practice problems and help, check out our Practice.

Until next time,

Leah