Mastering Factoring: How to Factor (x+3)^3 with Ease

[rcmango]

how to factor (x+3)^3

maybe foil after factoring?

thanks for any help.

 

[CRGreathouse]

Huh? It's already factored.

 

[rcmango]

hmm, maybe i used the wrong terminolgy, i want to take it apart, is that possible?

is (x+3)^3 the same as (x+3)(x+3)^2

 

[Hurkyl]

Ah, you want to expand it. (There are other synonyms too)

Yes, those two are the same; that's essentially the definition of raising something to the third power.


Factoring is the process where, for example, you start with

x^2 + 4x + 3

and end up with

(x+3)(x+1).

 

[rcmango]

okay so to fully expand (x+3)^3

could i just take the foil of (x+3)^2 and then multiply by (x+3) again?

if so, I'm unsure how to multiply the x+3 and the foiled polynomial

 

[Parthalan]

(a+b)^2 = a^2 + 2ab + b^2
(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3

There's a really easy trick to finding these. Expanding expressions of the form (x+y)^n can be done as follows:

First write x^n. The coefficient of the next term is equal to the current coefficient multiplied by the current exponent of x, and divided by the term number. For the coefficient of the second term, this is 1 * n / 1 = n. Decrease the exponent of x by 1, and increase the exponent of y by 1. Repeat this process until you get to y^n.

If they are of the form (x-y)^n, the signs just alternate. If you're unsure, use the property that a - b = a + (-b).

Hope that helps!

 

[HallsofIvy]

"Foil" is just a way of remembering that you must multiply each term in one factor by each term in the other.

Using "foil" on (x+3)(x+3) give x²+ 3x+ 3x+ 9= x²+ 6x+ 9. Now to multiply that by x+3 again, first multiply each part by x: x(x²+ 6x+ 9)= x³+ 6x²+ 9x, then multiply each part by 3: 3(x²+ 6x+ 9)= 3x²+ 18x+ 27, and finally add: x³+ 6x²+ 9x+ 3x²+ 18x+ 27= x³+ 9x²+ 27x+ 27.

This is not "abstract algebra"- I'm moving it to "General Math".