Question

How can you quickly expand a binomial raised to a power?

Answer 1

Binomial Theorem and Pascals Triangle

Explanation:

You can use pascals triangle to find the coefficient and binomial theorem to find the exponents.

Binomial Theorem

Take for example the general binomial:
`(a+b)^n`, from here we can expand this out just by using binomial theorem at first which essentially says "a" will be raised to the power of "n" and b will be raised to the power of 0 at first, but then as you go to each next term, the exponent of "a" decreases by 1, while the exponent of "b" increases by 1, until the exponent of "b" reaches "n" and the exponent of "a" reaches 0.

This can be illustrated as:
`C_0a^nb^0+C_1a^(n-1)b^1+C_2a^(n-2)b^2...C_na^0b^n`, while this make look intimidating at first, the C's are just the coefficient (which we don't know yet), we only know the exponent using Binomial Theorem. Essentially as one decreases the other increases until you have a total of "n+1" terms in the expression (which is when the exponent of b equals "n")

Pascals Triangle

Pascals triangle starts with one value "1" at the top, and this is considered "row 0" and then from here every row below this is defined as the sum of the two upper terms. I attached an image to illustrate this.

This triangle is super useful because it gives us the coefficients that we previously didn't know. They correspond to the expression in their respective order.

Example Problem

I'll do a fairly simply problem, but one that would take a bit without pascals triangle and binomial theorem and would involve a lot of terms.

`(x+4)^4`

Let's just start by using Binomial Theorem:
`C_0x^4*4^0+C_1x^3*4^1+C_2x^2*4^2+C_3*x^1*4^3+C_4x^0*4^4`

This simplifies to:
`C_0x^4 + C_1x^3*4 + C_2x^2 * 16 + C_3x * 64 + C_4 * 256`

Now from here we can find the coefficients using pascals triangle. I anote: the first row is considered row 0)

In the 4th row we get the following values: 1, 4, 6, 4, 1 which are going to correspond to the coefficients in their respective order, which just means:
`C_0=1, C_1=4, C_2=6, C_3=4, C_4=1`

plugging these values in for the coefficients we get

`x^4 +16x^3+96x^2+256x+256`

while this process may look a bit long, it's a lot easier than the traditional method of distributing the terms which you would need to do 4 times and would have a lot of terms taking a lot longer.