## 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