how do you factor?

Question

x2-2xy-3y2

Answer 1

you can use trial and error, but this is a way my teacher taught me.

x2 - 2xy - 3y2

N1 + N2 = -2 (the number in the middle)
N1 * N2 = -3 (the first and the third numbers multiplied together)

find N1 and N2... what added together is -2 and multiplied together is -3? (answer: -3 and 1)
so...

x2 + (1)xy - 3xy - 3y2

now, group the two parts together.

(x2 + xy) + (- 3xy - 3y2)

now take the common parts out of both groups.

(x)(x + y) + (-3y)(x + y)

now you see that both parentheses have the same things. this is going to be one parentheses of your final answer. the other parentheses will be the two common parts taken out of the groups. so it will look like this:

(x + y)(x - 3y)

combined, like this:
x2 - 2xy - 3y2
= x2 + xy - 3xy - 3y2
= x(x + y) - 3y (x + y)
= (x + y)(x - 3y)

hope this wasn't too confusing!

Answer 2

Note that you have the form
a^2 ± bxy ± cy^2
This implies a possible solution of
(x + my)(x + ny)
expanding this, we get
x^2 + (m+n)xy + mny^2, so
m + n ≡ -2, and
mn ≡ -3
-3 has only 2 possible real, integer factors:
-1, 3 and
1, -3
-1 + 3 = + 2, so that can't be a solution.
1 + -3 = -2, which is the solutiom, so
x2-2xy-3y^2 =
(x + 1)(x - 3)

Answer 3

Use trial and error as suggested. Or read down and follow the decomposition method someone showed on an earlier question.

Answer 4

You need know 1. a^2 - b^2 = (a+b) (a-b) = a^2 - ab + ab - b^2 = a^2 - b^2 and 2. ( a - b)^2 = a^2 - 2ab + b^2 and you don't have problem with it.

Answer 5

x^2 - 2xy -3y^2 will factor as follows: (x + y)(x - 3y). I leave it to you to check.

Answer 6

add the x's

and the y's

Answer 7

(x+y)(x-3y)

Just trial and error.