I'm not sure why this is labeled as calculus; it's just algebra.
You take the first coefficient and the 2nd coefficient and add them.
1y x 3y = (1+3)y = 4y
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4xy²(3y-x)-2x²(x-3y)²=4xy²(x-3y)-2x²(x-3y)²=(x-3y)[4xy²-2x²(x-3y)²]=(x-3y)(4xy²-2x³+6x²y)=(x-3y)(-2x³+6x²y+4xy²)=-2x(x-3y)(x²-3xy-2y²)= -2x(x-3y)(x-y)(x-2y)
3xy + 3y2 = 3y (x + y)
43
3y2 y x y = y2 3 x y2 = 3y2
X + 3y - 2y + 3y = X + (3-2+3)y = X + 4y