how do u factor a expressions like x^2-6x or 15x^3y-35x^2y^2+40xy^3?

Question

Alg 2
Factoring Out a Monomial

Answer 1

A monomial is something like:

2
x
2x

So look at your first expression: what is common to both terms? The x. So, divide (factor out) the x, what do you have left?

x(x-6)

Now your other question is the same thing, it just has more terms:

15x^3 y - 35x^2 y^2 + 40xy^3

The first thing I notice is that the coefficients are all multiples of 5. So we can factor out 5 for sure. But what about all those x's and y's? How many are common to each term? They each have an x, they each have a y. But they don't all have x^2 or anything like that. So we'll factor out 5xy:

5x(3x^2 - 7xy + 40y^2)


Another way to look at it is to write out all the factors of the terms and cross out or circle what's in common.

x^2= x * x
-6x = -6 * x

They both have x in common, so that's one factor, what's left is x-6.

Same thing can be done for the other one:
15x^3 y= 3 * 5 * x * x * x * y
-35x^2 y^2= -7 * 5 * x * x * y * y
40xy^3= 5 * 4 * 2 * x * y * y * y

If your division skills are good, the first way is fine. If you are getting lost with it and not clearly seeing what's in common, then take the longer process of writing out all the factors. You will find that eventually you won't need it.

Answer 2

look for what each term has in common:

x^2 - 6x
each term has an x so
x (x - 6)

15 x^3 y - 35 x^2 y^2 + 40 x y^3
each term has a factor of 5, an x and a y so
5xy (3 x^2 - 7xy + 8 y^2)
that's as far as that one simplifies

Answer 3

First, pull out an x for any that has an x in each expression. First one would be x(x-6). That would be the factors of it. Same could be done for that second expression, but you could pull out an x and y. 5xy(3x^2-7xy+8y^2).

Answer 4

tell me the anser to that
when you get it