Algebra Help?

Question

I am having trouble figuring out how to factor problems so please help.

Example: 10xy-20x² y+3x-6x²

Please explain in detail because I am not good with algebra at all.

Answer 1

10xy-20x² y+3x-6x²
=x(10y -20xy+3 -6x)

Answer 2

10xy - 20x^2 y + 3x - 6x^2

First look for any factor all terms have in common

x(10y - 20xy + 3 - 6x)

The expression in ( ) can be factored by grouping

Factor 10y from the first two terms and 3 from the last two terms

x[10y(1 - 2x) + 3(1 -2 x)]

The expression in [ ] is, believe it or not, of the same form as the right side of the distributive rule

a(b + c) = ab + ac
Here a = (1 - 2x), b = 10y, c = 3

x(1 - 2x)(10y - 3)

Remember that 10y(1 - 2x) = (1 - 2x)(10y)

Answer 3

The expression is:
10xy -20x^2y + 3x - 6x^2
10xy(1 - 2x) + 3x((1 -2x)
(1 -2x)(10xy + 3x)
x(1 - 2x)(10y + 3+

Answer 4

You just look first for common factors, the same factors in each term.

10xy has factors 2*5*x*y
20x²y is 2*2*5*x*x*y
3x is just 3*x
6x² is 2*3*x*x

The only factor they all have in common is x (just once).

Answer 5

Hey my name is Christa... First extend all exponents...so it would be 10xxy-20xxy+3x-6xx, Then change all #s after sub. sign to a negitive and the sub. sign to add, so it would be 10xxy+(-20)xxy+3x+(-6)xx, Then combine(add) all like terms so 10+(-20)+3+(-6) which = -13, Then add all like varibles so x+x+x+x+x+x+x= x to 7th power and then y+y= y to the 2nd power and your answer would be -13x(to the 7th power) y (to the 2nd power). I'm in 6th grade but in the gem class ( whichs means I do 8th grade math) what grade are you in? : )