The expression \begin{align*}(2x+3)^5\end{align*} would take a while to multiply out. Is there a pattern you can use?

#### Watch This

http://www.youtube.com/watch?v=NLQmQGA4a3M James Sousa: The Binomial Theorem Using Pascal’s Triangle

#### Guidance

Pascal was a French mathematician in the @$\begin{align*}17^{th}\end{align*}@$ century, but the triangle now named Pascal’s Triangle was studied long before Pascal used it. The pattern was used around the @$\begin{align*}10^{th}\end{align*}@$ century in Persia, India and China as well as many other places.

The primary purpose for using this triangle is to introduce how to expand binomials.

@$$\begin{align*}(x+y)^0 &= 1\\ (x+y)^1 &= x+y\\ (x+y)^2 &= x^2+2y+y^2\\ (x+y)^3 &= x^3+3x^2y+3xy^2+y^3\end{align*}@$$

Notice that the coefficients for the @$\begin{align*}x\end{align*}@$ and @$\begin{align*}y\end{align*}@$ terms on the right hand side line up exactly with the numbers from Pascal’s triangle. This means that given @$\begin{align*}(x+y)^n\end{align*}@$ for any power @$\begin{align*}n\end{align*}@$ you can write out the expansion using the coefficients from the triangle. When you study how to count with combinations then you will be able to calculate the value of any coefficient without writing out the whole triangle.

There are many patterns in the triangle. Here are just a few.

- Notice the way each number is created by summing the two numbers above on the left and right hand side.
- As you go further down the triangle the values in a row approach a bell curve. This is closely related to the normal distribution in statistics.
- For any row that has a second term that is prime, all the numbers besides 1 in that row are divisible by that prime number.
- In the game Plinko where an object is dropped through a triangular array of pegs, the probability (which corresponds proportionally to the values in the triangle) of landing towards the center is greater than landing towards the edge. This is because every number in the triangle indicates the number of ways a falling object can get to that space through the preceding numbers.

**Example A**

Expand the following binomial using Pascal’s Triangle: @$\begin{align*}(3x-2)^4\end{align*}@$

**Solution: ** The coefficients will be 1, 4, 6, 4, 1; however, since there are already coefficients with the @$\begin{align*}x\end{align*}@$ and the constant term you must be particularly careful.

@$\begin{align*}1 \cdot (3x)^4+4 \cdot (3x)^3 \cdot (-2)+6 \cdot (3x)^2 \cdot (-2)^2+4 \cdot (3x) \cdot (-2)^3+1 \cdot (-2)^4\end{align*}@$

Then it is only a matter of multiplying out and keeping track of negative signs.

@$\begin{align*}81x^4-216x^3+216x^2-96x+16\end{align*}@$

**Example B**

Expand the following trinomial: @$\begin{align*}(x+y+z)^4\end{align*}@$

**Solution: ** Unfortunately, Pascal’s triangle does not apply to trinomials. Instead of thinking of a two dimensional triangle, you would need to calculate a three dimensional pyramid which is called Pascal’s Pyramid. The sum of all the terms below is your answer.

@$$\begin{align*}&1x^4+4x^3z+6x^2z^2+4xz^3+1z^4\\ &4x^3y+12x^2yz+12xyz^2+4yz^3\\ &6x^2y^2+12xy^2z+6y^2z^2\\ &4xy^3+4y^3z\\ &1y^4\end{align*}@$$

Notice how many patterns exist in the coefficients of this layer of the pyramid.

**Example C**

Expand the following binomial: @$\begin{align*}\left(\frac{1}{2}x-3\right)^5\end{align*}@$

**Solution: ** You know that the coefficients will be 1, 5, 10, 10, 5, 1.

@$$\begin{align*}1 \left(\frac{1}{2}x\right)^5+5 \left(\frac{1}{2}x\right)^4 (-3) +10 \left(\frac{1}{2}x\right)^3 (-3)^2+10 \left(\frac{1}{2}x\right)^2 (-3)^3+5 \left(\frac{1}{2}x\right) (-3)^4+1 \cdot (-3)^5\end{align*}@$$

@$$\begin{align*}=\frac{x^5}{32}-\frac{15x^4}{16}+\frac{90x^3}{8}-\frac{270x^2}{4}+\frac{405x}{2}-243\end{align*}@$$

Remember to simplify fractions.

@$$\begin{align*}=\frac{x^5}{32}-\frac{15x^4}{16}+\frac{45x^3}{4}-\frac{135x^2}{2}+\frac{405x}{2}-243\end{align*}@$$

**Concept Problem Revisited**

Pascal’s triangle allows you to identify that the coefficients of @$\begin{align*}(2x+3)^5\end{align*}@$ will be 1, 5, 10, 10, 5, 1 like in Example C. By carefully substituting, the expansion will be:

@$$\begin{align*}1 \cdot (2x)^5+5 \cdot (2x)^4 \cdot 3 + 10 \cdot (2x)^3 \cdot 3^2+10 \cdot (2x^2) \cdot 3^3+5(2x)^1 \cdot 3^4+3^5\end{align*}@$$

Simplifying is a matter of arithmetic, but most of the work is done thanks to the patterns of Pascal’s Triangle.

#### Vocabulary

A ** binomial expansion** is a polynomial that can be factored as the power of a binomial.

** Pascal’s Triangle** is a triangular array of numbers that describes the coefficients in a binomial expansion.

#### Guided Practice

1. Factor the following polynomial by recognizing the coefficients.

@$\begin{align*}x^4+4x^3+6x^2+4x+1\end{align*}@$

2. Factor the following polynomial by recognizing the coefficients.

@$\begin{align*}8x^3-12x^2+6x-1\end{align*}@$

3. Expand the following binomial using Pascal’s Triangle.

@$\begin{align*}(A-B)^6\end{align*}@$

**Answers:**

1. @$\begin{align*}(x+1)^4\end{align*}@$

2. Notice that the first term of the binomial must be @$\begin{align*}2x\end{align*}@$, the last term must be -1 and the power must be 3. Now all that remains is to check.

@$\begin{align*}(2x-1)^3=(2x)^3+3(2x)^2 \cdot (-1)+3(2x)^1(-1)^2+(-1)^3=8x^3-12x^2+6x-1\end{align*}@$

3. @$\begin{align*}(A-B)^6=A^6-6A^5B+15A^4B^2-20A^3B^3+15A^2B^4-6AB^5+B^6\end{align*}@$

#### Practice

Factor the following polynomials by recognizing the coefficients.

1. @$\begin{align*}x^2+2xy+y^2\end{align*}@$

2. @$\begin{align*}x^3+3x^2+3x+1\end{align*}@$

3. @$\begin{align*}x^5+5x^4+10x^3+10x^2+5x+1\end{align*}@$

4. @$\begin{align*}27x^3-27x^2+9x-1\end{align*}@$

5. @$\begin{align*}x^3+12x^2+48x+64\end{align*}@$

Expand the following binomials using Pascal’s Triangle.

6. @$\begin{align*}(2x-3)^3\end{align*}@$

7. @$\begin{align*}(3x+4)^4\end{align*}@$

8. @$\begin{align*}(x-y)^7\end{align*}@$

9. @$\begin{align*}(a+b)^{10}\end{align*}@$

10. @$\begin{align*}(2x+5)^5\end{align*}@$

11. @$\begin{align*}(4x-1)^4\end{align*}@$

12. @$\begin{align*}(5x+2)^3\end{align*}@$

13. @$\begin{align*}(x+y)^6\end{align*}@$

14. @$\begin{align*}(3x+2y)^3\end{align*}@$

15. @$\begin{align*}(5x-2y)^4\end{align*}@$