4x^2+2xy-3y^2-2x^2+5xy-2y2Collect the like terms=2x^2+7xy-5y^2We cannot factorise it because the end term is a negative whilst the second-last term is positive.
2(x - 1)(2x + y)
2xy + 4x + 8y + 16 = 2[xy + 2x + 4y + 8] = 2[x(y + 2) + 4(y + 2)] = 2(y + 2)(x + 4)
2xy=2*30*2=120
2 plus 1-x divided by.
( 2xy - 4x ) + ( 8y - 16)2 ( xy - 2x ) + 2 ( 4y - 8 ) Answer
2xy - 4x + 8y - 16 can be factored.(2xy - 4x) + (8y - 16)2x(y - 2) + 8(y - 2)Since y-2 is the same as the other, it would turn out like this:(2x + 8)(y - 2):]
(2x + 2y) is exactly the same thing as 2(x+y). In order to know how much it actually is, you have to know the numerical values of both 'x' and 'y'.
2 (2x2 + 2x + xy + y ) it's 2 (2x + y)(x + 1) if you're doing A+
4x^2+2xy-3y^2-2x^2+5xy-2y2Collect the like terms=2x^2+7xy-5y^2We cannot factorise it because the end term is a negative whilst the second-last term is positive.
2(x - 1)(2x + y)
2xy + 4x + 8y + 16 = 2[xy + 2x + 4y + 8] = 2[x(y + 2) + 4(y + 2)] = 2(y + 2)(x + 4)
8x-7 plus 2x equals 6 plus 5x plus 2?
2xy=2*30*2=120
2x + 5 = 13 2x = 8 2x/2 = 8/2 x=4
2 plus 1-x divided by.
2x+2=x+5 x=3