## Tuesday, October 31, 2017

### 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.

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.

(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

(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

## Wednesday, October 11, 2017

### Middle School Math Solutions – Expand Calculator, FOIL Method

In our last blog post we covered the distributive law. In this blog post, we will focus on an application of the distributive law, the FOIL method.

The FOIL method is used when multiplying two binomials together. Quick reminder: a binomial is an expression of the sum or difference of two terms.

The FOIL method is when we take the sum of the first two terms multiplied together, the outer terms multiplied together, the inner terms multiplied together, and the last terms multiplied together. This is where we get the acronym FOIL (first, outer, inner last).

Let’s visualize this and see it in a formula:

Let’s see some examples now, using the FOIL method.

(x+3)(x+1)

1.  Use the FOIL method

Remember, we want the sum of the first terms multiplied together, the outer terms multiplied together, the inner terms multiplied together, and the last terms multiplied together.
2.  Simplify

(x+3)(x+1)=x∙x+x∙1+3∙x+3∙1

=x^2+x+3x+3

=x^2+4x+3

Multiply terms and add like terms

Not, too complicated. Let’s see two more examples.

(2x+1)(x-1)

1. Use the FOIL method

2. Simplify

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

=2x^2-x-1

(3x^2+y)(-x+5y)

1. Use the FOIL method

2. Simplify

(3x^2+y)(-x+5y)=-3x^3+15x^2 y-xy+5y^2

As you can see, multiplying binomials using the FOIL method isn’t hard. The FOIL method is so helpful throughout your math courses, so it is important to memorize and get the hang of it. Once you start practicing, this will become second nature to you. For practice problems or more help on the FOIL method, check out our Practice.

Until next time,

Leah