## Tuesday, January 16, 2018

### Middle School Math Solutions – Order of Operations Calculator

Given a problem like, 3×5-2, you might be wondering “how do I solve this?” since there are two possible ways.

Option 1: (3×5)-2=15-2=13

Option 2: 3×(5-2)=3×3=9

This is where the order of operations comes in, since there is only one correct answer.

The order of operations are rules that tell us which operation to perform first when calculating an algebraic expression.

Here is the order of operations:

1.  Parentheses
2.  Exponents
3.  Multiplication and Division (from left to right)
4.  Addition and Subtraction (from left to right)

When calculating an algebraic expression, go down the order of operations, starting with parentheses, and perform the operations that apply.

There is an acronym used to remember the order of operations, PEMDAS (parentheses, exponents, multiplication, division, subtraction, and addition). The phrase “Please Excuse My Dear Aunt Sally” is also used to help remember the order by using the first letter of each word.

Looking back at the problem above and knowing the order of operations, we now can see that the answer is 13, option 1.

Let’s see some examples using the order of operations.

8×5-(6+10)÷2

1. Calculate within parentheses

=8×5-16÷2
We simplified what was in the parentheses.
2. Multiplication and Division (from left to right)

=40-16÷2
=40-8
We multiplied and divided starting from the left of the expression, making our way to the right.
3. Addition and Subtraction (from left to right)

=40-8
=32

2×(5×6+2^2 )+(8÷2)

1. Calculate within parentheses – Evaluate (5×6+2^2)

(5×6+2^2 )=(5×6+4)
=(30+4)
=34
Make sure you apply the order of operations when calculating inside the parentheses
Now we have: 2×(5×6+2^2 )+(8÷2)=2×34+(8÷2)
2. Calculate within parentheses – Evaluate  (8÷2)

8÷2=4
Now we have: 2×34+(8÷2)=2×34+4
3. Multiplication and Division (from left to right)

=2×34+4
=68+4

4. Addition and Subtraction (from left to right)

=68+4
=72

6+(2(4+2))^2-5^2

1. Calculate within parentheses

=6+(2(6) )^2-5^2
=6+12^2-5^2

2. Exponents

=6+144-5^2
=6+144-25

3. Addition and Subtraction (from left to right)

=150-25
=125

Learning the order of operations is essential in math because they will guide you to the correct answer. For more help or practice with the order of operations, check out Symbolab’s Practice.

Until next time,

Leah

## 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.

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).

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

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