(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).
It's m = y2 - y1/ x2- x1 It's m equals y2 minus y1 over x2 minus x1
x^3 - 121x
For any x ≠ 0, x2 -10/x2 - 4 +3x/x2 - 4 LCD = x2, multiply each term by their missing element of LCD = (x4 + 10 +3x - 8x2)/x = (x4 - 8x2 + 3x + 10)/x2
2x2=4 X -2=-8
1 over x2 - 4 is the multiplicative inverse of x2 minus four 1/x2 - 4
x2-3x-10 = (x+2)(x-5) when factored
(x2 + 1)(x2 - 2)
x3-x2
x2 - (2x)2 = x2 - 4x2 = -3x2 x2 - 2x2 = -x2
x2 - 36 = x2 - 62 = (x + 6)(x - 6)
x4 - x3 - x - 1 rewriting: = x4 - 1 - x3 - x factorising pair of terms: =(x2 + 1)*(x2 - 1) - x*(x2 + 1) = (x2 + 1)*(x2 - 1 - x) or (x2 + 1)*(x2 - x - 1) which cannot be factorised further.
(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).
y2(minus)y1 ---------------(over) x2(minus)x1
It's m = y2 - y1/ x2- x1 It's m equals y2 minus y1 over x2 minus x1
x3 + x2 - 6x + 4 = (x - 1)(x2 + 2x - 4)
x^3 - 121x