Wednesday, January 3, 2018

Middle School Math Solutions – Expand Calculator, Binomial Expansion

We’ve learned how to expand perfect squares and perfect cubes. Now, we are going to learn how to expand binomials raised to any positive integers.

Imagine having to expand a binomial raised to a power of 7. Sounds like a lot of work, right? Good news, we have a formula!

Here is the Binomial Theorem:

                                            (a+b)^n=\sum_{i=0}^n\binom{n}{i}a^(n-i)b^i

\binom{n}{i} is a combination, which we read as “n choose i”.
Here is the formula for n choose i:

                                                        \binom{n}{i}=\frac{n!}{i!(n-i)!}

Let’s see some examples using this formula.

First example (click here):

                                                          Expand (x+2)^4

1. Apply the formula

                                                         a=x, b=2, n=4

                                            (x+2)^4=\sum_{i=0}^4\binom{4}{i}x^(4-i)2^i

2. Expand the summation and simplify

           \sum_{i=0}^4\binom{4}{i}x^(4-i)2^i =\binom{4}{0} x^4∙2^0+\binom{4}{1} x^3∙2^1+\binom{4}{2} x^2∙2^2+\binom{4}{3} x^1 ∙2^3+\binom{4}{4} x^0∙2^4

                                       =\frac{4!}{0!(4-0)!} x^4∙2^0+\frac{4!}{1!(4-1)!} x^3∙2^1 
                   
                                          +\frac{4!}{2!(4-2)!} x^2∙2^2+\frac{4!}{3!(4-3)!} x^1 ∙2^3+\frac{4!}{4!(4-4)!} x^0∙2^4
           
                                 =\frac{24}{(1)24} x^4∙2^0+\frac{24}{1(6)} x^3∙2^1+\frac{24}{2(2)}x^2∙2^2+\frac{24}{6(1)} x^1 ∙2^3+\frac{24}{24(1)} x^0∙2^4

                                      =x^4+4x^3∙2+6x^2∙4+4x∙8+1∙16
                      
                                      =x^4+8x^3+24x^2+32x+16 
                                   
You can see that there is a lot to calculate, but that this formula makes expanding easier and faster. In this example, I went into detail on how to simplify this expansion, specifically the combination. Now, that you’ve seen and understand how to calculate combinations, the next examples won’t be in such detail (you can also check out a more detailed step by step solution by clicking the hyperlinks).

Next example (click here):

                                                             Expand (x-y)^5

1. Apply the formula

                                                              a=x, b=-y, n=5

                                           (x-y)^5=\sum_{i=0}^5\binom{5}{i} x^(5-i) (-y)^i

2. Expand the summation and simplify

               \sum_{i=0}^5\binom{5}{i} x^(5-i) (-y)^i
                                  =\binom{5}{0} x^5∙(-y)^0+\binom{5}{1} x^4∙(-y)^1+\binom{5}{2} x^3∙(-y)^2+\binom{5}{3} x^2 ∙(-y)^3+\binom{5}{4} x^1∙(-y)^4+\binom{5}{5} x^0 (-y)^5
               =x^5-5x^4 y+10x^3 y^2-10x^2 y^3+5xy^4-y^5

Last example (click here):

                                                          Expand (3+x^2 )^4

1. Apply the formula

                                                          a=3, b=x^2, n=4

                                           (3+x^2)^4=\sum_{i=0}^4\binom{4}{i} 3^(4-i) (x^2)^i

2. Expand the summation and simplify

            \sum_{i=0}^4\binom{4}{i} 3^(4-i) (x^2 )^i
                                                                          =\binom{4}{0} 3^4∙(x^2)^0+\binom{4}{1} 3^3∙(x^2 )^1+\binom{4}{2} 3^2∙(x^2)^2+\binom{4}{3} 3^1 ∙(x^2 )^3+\binom{4}{4} 3^0∙(x^2 )^4

                     =81+4∙27x^2+6∙9x^4+4∙3x^6+x^8
                                    
                     =81+108x^2+54x^4+12x^6+x^8
                                            
Binomial expansions require practice to get the hang of things and to help memorize the formula. If you are interested in more practice problems on this topic or help, check out Symbolab’s Practice.

Until next time,

Leah

Wednesday, December 20, 2017

Middle School Math Solutions – Expand Calculator, Perfect Cube

Similarly to the perfect square formula, which we covered in the last post, we get the perfect cube formula. The perfect cube is when we cube a binomial. We use polynomial multiplication and the FOIL method for this formula.

Let’s take a look at it:

                                                    (a+b)^3=a^3+3a^2 b+3ab^2+b^3

Here’s the proof:

                                                    (a+b)^3=(a+b)(a+b)(a+b)
                                                                   =(a+b)(a^2+2ab+b^2)
                                                                   =a(a^2+2ab+b^2 )+b(a^2+2ab+b^2)
                                                                   =a^3+2a^2 b+ab^2+a^2 b+2ab^2+b^3
                                                                   =a^3+3a^2 b+3ab^2+b^3

As you can see there are many steps involved, so having this formula will help you solve perfect cube problems quickly.

Let’s see some examples.

First example (click here):

                                                                  Expand (x+1)^3

 1.  Apply the formula

                                                                       a=x,b=1

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

Next example (click here):

                                                                 Expand (2x-2)^3

1. Apply the formula

                                                                    a=2x,b=-2

                               (2x-2)^3=(2x)^3+3∙(2x)^2∙(-2)+3∙(2x)∙(-2)^2+(-2)^3
                                                =8x^3-24x^2+24x-8                             

Last example (click here):

                                                               Expand (s^2+2t)^3

1. Apply the formula

                                                                   a=s^2,b=2t

                                 (s^2+2t)^3=(s^2 )^3+3∙(s^2 )^2∙(2t)+3∙s^2∙(2t)^2+(2t)^3
                                                   =s^6+6s^4 t+12s^2 t^2+8t^3               

Not too difficult! Memorizing this formula will help you and become faster at cubing binomials. If you are looking for more practice problems or help, check out Symbolab’s Practice.

Until next time,

Leah

Wednesday, November 15, 2017

Middle School Math Solutions – Expand Calculator, Perfect Squares

The perfect square formula is an application of the FOIL method that will help you calculate the square of a binomial quickly.

Let’s take a look at the formula:

                                                     (a+b)^2=a^2+2ab+b^2

Here’s the proof:

                                                    (a+b)^2=(a+b)(a+b)
                                                                   =a^2+ab+ab+b^2
                                                                   =a^2+2ab+b^2

Now, let’s see some examples using the perfect squares formula.

First example (click here):

                                                         Expand (x+1)^2

1. Apply the formula

                                                              a=xb=1

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

Second example (click here):

                                                         Expand (5-x)^2

1. Apply the formula

                                                             a=5b=-x

                                             (5-x)^2=5^2-2∙5∙(-x)+(-x)^2
                                                            =25+10x+x^2

Last example (click here):

                                                      Expand (s^2+4p)^2

1. Apply the formula

                                                            a=s^2, b=4p

                                        (s^2+4p)^2=(s^2)^2+2∙s^2∙4p+(4p)^2
                                                           =s^4+8s^2 p+16p^2

The more you practice these problems, the faster you will be able to do them. I’ve used this formula so much, that now, I don’t need it since I can do it all mentally. For more help or practice on this topic, check out Symbolab’s Practice.

Until next time,

Leah

Tuesday, November 7, 2017

Middle School Math Solutions – Expand Calculator, Two Squares

The difference of two squares is an application of the FOIL method (refer to our blog post on the FOIL method).   The difference of two squares is a number or term squared subtracted from another number or term squared. We get it when we multiply two binomials, where the terms in the binomials are the same, except one of the terms is subtracted instead of being added.

Let’s see this in a formula:
                                                             (a+b)(a-b)=a^2-b^2

Here’s the proof using the FOIL method:
                                                      (a+b)(a-b)=a^2-ab+ab-b^2
                                                                                =a^2-b^2

Not too complicated, so let’s see some examples.

First example (click here):
                                                             Expand (x+2)(x-2)

1. Apply the formula
                                                                        a=x,b=2

                                                             (x+2)(x-2)=x^2-2^2
                                                                                       =x^2-4

This next one is a little more complicated.

Next example (click here):
                                                           Expand (-y+2x)(y+2x)

1. Rewrite the problem
                                                      (-y+2x)(y+2x)=(2x-y)(2x+y)
                                                                                      =(2x+y)(2x-y)

2. Apply the formula

                                                                    a=2x,b=y

                                                     (2x+y)(2x-y)=(2x)^2-y^2
                                                                                   =4x^2-y^2

For our last example, we will see an application of the difference of two squares formula.

Last example (click here):
                                                                        23∙17

1. Rewrite the numbers
                                                                (20+3)(20-3)

2. Apply the formula
                                                                     a=20,b=3

                                                          (20+3)(20-3)=20^2-3^2
                                                                                        =400-9
                                                                                        =391

As you can see, this formula is simple, but very helpful. If you need more help or practice with this formula, check out Symbolab’s Practice.

Until next time,

Leah

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.

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

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.

First example (click here):

                                                                 (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.

Next example (click here):

                                                                    (2x+1)(x-1)

1. Use the FOIL method


2. Simplify

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

                                                                     =2x^2-x-1

Last example (click here):

                                                                 (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

Monday, September 25, 2017

Middle School Math Solutions – Expand Calculator, Distributive Law

The distributive law helps with multiplication problems by breaking down large numbers into smaller numbers. In this blog post, we will talk about the distributive law and how to use it.

The law says that multiplying a number by a group of numbers added together is the same as multiplying each separately. What does this mean? Here we can see it in a formula:


You can see that the multiplication of a has been distributed among the sum of b and c.

Let’s see some examples of how to use the distributive law.

First example (click here):

                                                                Expand 4(2x+5)

1. Define the values for a, b, c

                                                               a=4,   b=2x,   c=5
Refer to the formula to see what these values are.
2. Plug these values into the distributive law formula

                                                               4(2x+5)=4∙2x+4∙5

3. Simplify your answer

                                                                4∙2x+4∙5=8x+20
8x+20 is our answer.
Next example (click here):

                                                                Expand 5(10-9p)

1. Define the values for a, b, c

                                                              a=5,   b=10,   c=-9p

2. Plug these values into the distributive law formula

                                                            5(10-9p)=5∙10+5∙(-9p)

3. Simplify your answer

                                                             5∙10+5∙(-9p)=50-45p
50-45p is the answer.
In this last example, we will see another application of the distributive law.

Last example (click here):

                                                                     5(106)

Just by looking at this problem, this might be difficult to calculate quickly. So let’s use the distributive law.

1. Make 106 the sum of two numbers

                                                                 106=100+6

2. Plug this in for 106

                                                                  5(100+6)

3. Define values a, b, c

                                                            a=5,   b=100,   c=6

4. Plug these into the distributive law formula

                                                           5(100+6)=5∙100+5∙6

5. Simplify your answer

                                                         5∙100+5∙6=500+30=530
530 is the answer.
As you can see, the distributive law is very handy. Therefore, I highly suggest you memorize this formula and become familiar with how to use it. You can do so by doing as many examples as you can. For practice examples and more help, check out Symbolab’s Practice

Until next time,

Leah