You can't solve this problem, since you don't know what yequals. So, the quotient is y over 4 cubed.
You could mean y/4 - 3, or you could possibly mean 4/y - 3. It is somewhat ambiguous which quotient you mean.
265
abs(y^3 - 4) < 8*y
The quotient works out as: x^2+2x+4 and there is a remainder of -3
9y cubed plus 2y squared
4x cubed y cubed z divided by x negative squared y negative 1 z sqaured = 4
You could mean y/4 - 3, or you could possibly mean 4/y - 3. It is somewhat ambiguous which quotient you mean.
265
abs(y^3 - 4) < 8*y
64 x cubed minus y cubed is (4x - y)(16x^2 + 4xy + y^2)
3
X / ( y + 4 )
(y + 27)(y^2 - 27y + 729)
The quotient works out as: x^2+2x+4 and there is a remainder of -3
It is x^2 + xy + y^2
To find two numbers that have a sum of 60 and a quotient of 4, let’s denote the numbers as ( x ) and ( y ), where ( x > y ). From the conditions, we have the equations ( x + y = 60 ) and ( \frac{x}{y} = 4 ). Solving these, we find ( x = 48 ) and ( y = 12 ). Thus, the two numbers are 48 and 12.
y^(3) - 2y^(2) + 4y Factor out 'y' Hence y(y^2 - 2y + 4) This will not factor any further.