Solve 2x^2 - 3xy + y^2?

Question

so lost!

Answer 1

2x^2 - 3xy^2

Answer 2

2x^2 - 3xy + y^2

write -3xy as -2xy - xy, it becomes like this

2x^2 - 2xy - xy + y^2

Now take 2x common from first 2 terms and -y from last 2 terms, it will become,

2x(x - y) - y(x - y)

Now take (x - y) common from the above equation, you will have it simplified as

(2x - y) (x - y)

Simplified.

Answer 3

i think it would be

-6x^3y^3

but im noy POSITIVE.  

Answer 4

u mean factor?

(2x-y)(x-y)

Answer 5

=(2x-y)(x-y)

Answer 6

(2x-y)(x-y)

Answer 7

When factoring, you need to reverse the FOIL process (First, Inside, Outside, Last).

Two factors will multiply to the starting expression.  Start with the first value (2x^2).  Two values that will multiply to this are 2x and x.  So....

(2x +  )(x +  )

Two values that will multiply to y^2 are y and y. So....

(2x + y)(x + y)

Multiply this out using the FOIL method...

2x^2 + yx + 2xy + y^2

2x^2 + 3xy + y^2

Note that the inside value is +3xy not -3xy.  You need to play with signs.

Try

(2x - y)(x - y)

2x^2 - yx - 2xy + y^2

2x^2 - 3xy + y^2

That is your starting expression, so the answer is...

(2x - y)(x - y)

------------

Note that

(y - 2x)(y - x) is also a valid solution

Answer 8

(2x-y)(x-y) is the factored form

If the equation is equal to 0, then x = (1/2)y Or x=y