The binomial theorem is a shortcut that we can use to expand expressions that have binomial bases raised to an exponent.

If you were to simplify a binomial power like\((2x-9)^7\) the long way, you would have to use polynomial multiplication to multiply \((2x-9)\) by itself 7 times. That would take a LONG time.

The binomial expansion process is much faster and it is pretty easy once you get the hang of it.

\[(x+y)^n= \sum_{k=0}^{n} {n \choose k}{x^{n-k}}{y^{k}}\]

The standard formula for the binomial theorem uses summation notation and combination notation.

If you haven't learned these notations yet, you can still use the binomial theorem by following the patterns shown in these binomial expansions:

\((x+y)^2=1x^2+2xy+1y^2\)

\((x+y)^3=1x^3+3x^2y+3xy^2+1y^3\)

\((x+y)^4=1x^4+4x^3y+6x^2y^2+4xy^3+1y^4\)

\((x+y)^5=1x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+1y^5\)

What patterns do you notice?

There are 4 important patterns that show how you can use the exponent of any binomial power to expand it into a polynomial.

When the binomial power has an exponent of n...

- There are n+1 terms in the expanded polynomial.
- The coefficients match the nth row of Pascal's Triangle
- The powers of x decrease from n to 0.
- The powers of y increase from 0 to n.

As an example, look at the expansion of \((x+y)^4\).

\((x+y)^4=\) \(1\)\(x^4\)\(y^0\) \(+\) \(4\)\(x^3\)\(y^1\) \(+\) \(6\)\(x^2\)\(y^2\) \(+\) \(4\)\(x^1\)\(y^3\) \(+\) \(1\)\(x^0\)\(y^4\)

The exponent of the binomial power is a 4. If we add 1 to that exponent, it tells us that there are 5 terms in the expanded polynomial.

The coefficients of those terms match the 4th row of Pascal's Triangle. And the powers of x decrease from 4 to 0 while the powers of y increase from 0 to 4.

The binomial theorem will expand any power as long as the base of the exponent is a binomial (a polynomial with exactly 2 terms).

- Determine the number of terms by adding 1 to the exponent.
- Use Pascal's Triangle to find the coefficients of each term.
- List the first term of the binomial with decreasing powers.
- List the second term of the binomial with increasing powers.
- Simplify each term by simplifying the exponents and multiplying the coefficients if there are multiple coefficients in each term.

In the expansions above, I expanded the binomial powers of \((x+y)\). But you can also use the binomial theorem to expand powers of more complicated binomials like \((3x+8y)\) or \((x^2-8)\) or \((7x^5-4x^2)\).

If the terms of the binomial have multiple parts (coefficients, variables, exponents, etc.), then you need to put them inside of parentheses when you list them with decreasing/increasing powers.

You also have to follow the power rule and the distributive property for exponents when you simplify the terms of the expanded polynomial.

Expand \((3x+2)^4\).

The exponent of this binomial power is a 4. So, that means that there will be 5 terms in the expanded polynomial.

Using the 4th row of Pascal's Triangle, I'll start by listing the binomial coefficients of these 5 terms.

\(1\_\_+4\_\_+6\_\_+4\_\_+1\_\_\)

Next, I will list the first term of the binomial (3x) with decreasing powers from left to right.

\(1\)\((3x)^4\)\(\_+4\)\((3x)^3\)\(\_+6\)\((3x)^2\)\(\_+4\)\((3x)^1\)\(\_+1\)\((3x)^0\)\(\_\)

Then I will list the second term of the binomial (2) with increasing powers from left to right.

\(1\)\((3x)^4\)\(2^0\)\(+4\)\((3x)^3\)\(2^1\)\(+6\)\((3x)^2\)\(2^2\)\(+4\)\((3x)^1\)\(2^3\)\(+1\)\((3x)^0\)\(2^4\)

Then I will simplify all the exponents.

\((3x)^4=81x^4\)

\((3x)^3=27x^3\)

\((3x)^2=9x^2\)

\((3x)^1=3x\)

\((3x)^0=1\)

\(2^0=1\)

\(2^1=2\)

\(2^2=4\)

\(2^3=8\)

\(2^4=16\)

\(1\)\((81x^4)\)\((1)\)\(+4\)\((27x^3)\)\((2)\)\(+6\)\((9x^2)\)\((4)\)\(+4\)\((3x)\)\((8)\)\(+1\)\((1)\)\((16)\)

And then multiply the coefficients in each term.

\(1\)\((81x^4)\)\((1)\)\(=81x^4\)

\(4\)\((27x^3)\)\((2)\)\(=216x^3\)

\(6\)\((9x^2)\)\((4)\)\(=216x^2\)

\(4\)\((3x)\)\((8)\)\(=96x\)

\(1\)\((1)\)\((16)\)\(=16\)

Answer:

\((3x+2)^4\)

\(=\)

\(81x^4+216x^3y+216x^2y^2+96x+16\)

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