Factorising some equations?

Question

Hello,
Im stuck factorising some equations
i can do this sort of equation
x^2 + 4x + 3
=(x+1)(x+3)

but theres a couple i don't understand such as
x^2 – 2xy + 3y2
9p^2 – 4q^2
121p^2 – 64q^2

Any websites or help would be much apreciated
Cheers

Answer 1

x^2 - 2xy + 3y^2
= x^2 - 2xy + y^2 + 2y^2
= (x - y)^2 + 2y^2

9p^2 - 4q^2
= (3p)^2 - (2q)^2
= (3p-2q)(3p+2q)

121p^2 - 64q^2
=(11p)^2 - (8q)^2
= (11p-8q)(11p+8q)

Answer 2

In such equations you have to use the formula a^2-b^2 = (a+b) (a-b)

9p^2 - 4q^2 = (3p)^2 - (2q)^2
= (3p+2q) (3p-2q)

Similarly the next one;
(11p)^2 - (8q)^2 = (11p+8q) (11p-8q)

Answer 3

Don't be dependent on websites for solving eqn. They are very easy if you just try.
You can solve eqn in one variable, similarly try for 2 variables. Instead of numbers you just use an extra variable.
Your first eqn seems incorrect. Anyway, you can write
x^2 - 2xy + 3y^2 = (x - y)(x - 3y) + 2y

Use a^2 - b^2 = (a + b)(a - b)
For rest two.

Answer 4

I can try to help you if you like.

9p^2-4q^2
(3p-2q)(3p+2q)--the-6pq and 6pq cancel out

121p^2-64q^2
(11p-8q) (11p+8q)th2 19pq and -19pq cancel out

x^2-2xy+3y^2
is factored out already.

Answer 5

(1). (x+y)(x-3y)
(2). (3p+2q)(3p-2q)
(3). (11p+8q)(11p-8p)

Answer 6

1.(x-y)(x-y)
2.(3p+2q)(3p-2q)
3.(11p+8q)(11p-8q)