x2 / x = x
(x2-x-12)/(x-4) = (x+3)
x2 + 5x + 25
Quotient: 2x3-x2-14x+42 Remainder: -131 over (x+3)
2x / (x2) = 2*x / (x*x); cancel an x from numerator & denominator: 2 / x
-3
Z = 3x1+5x2+4x3 Subject to constraints2 x1 + 3x2 =8 3x1 + 2x 2 + 4x3=15 2x2 + 5x3 = 10 x1,x2,x3, =0
The remainder is 8. (x2 + 4)/(x - 2) = (x + 2) + 8/(x - 2) or x2 + 4 = (x - 2)(x + 2) + 8
Given the limited information in the question, Z is maximised when x1 or x2 (or both) are maximised. There is no trade-off between x1 and x2 to worry about.
5
(3x4 + 2x3 - x2 - x - 6)/(x2 + 1)= 3x2 + 2x - 4 + (-3x - 2)/(x2 + 1)= 3x2 + 2x - 4 - (3x + 2)/(x2 + 1)where the quotient is 3x2 + 2x - 4 and the remainder is -(3x + 2).
-3
Divide both sides of the equation by x: x2 = 9x x2 / x = 9x / x x = 9
x2 / x = x
(x2-x-12)/(x-4) = (x+3)
x2 + 5x + 25
x2 - 3X - 10 = (X - 5)(X + 2)So, (X2 - 3X -10) / (X-5) = X + 2