3 - 3x + x2 - x3 = (1 - x)(x2 + 3)
(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).
x2 + x + 6
x^2 + 3x - 2; -x^2 - 3x + 2andx - 1; 1 - x
(f - g)(x) = f(x) - g(g) = (3x + 1) - (x2 - 6) = -x^2 + 3x + 7
x2 + 3x + 7 = 5 ∴ x2 + 3x + 2 = 0 ∴ (x + 1)(x + 2) = 0 ∴ x ∈ {-2, -1}
X2+3x+x+3=x(x+1)+3(x+1)=(x+3)(x+1)
3 - 3x + x2 - x3 = (1 - x)(x2 + 3)
3(x2 + 3x + 1)
x2 + 3x + 2 factors into (x + 1) (x + 2)
3x-1 = 11 3x = 11+1 3x = 12 x = 4 Therefore: x2+2 = 18
x2 + 3px + p = 0 x2 + p(3x + 1) = 0 p(3x+1) = -x2 p = -x2/(3x+1) So p can have any value at all. In fact, around x = -1/3, p goes asymptotically to + and - infinity.
(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).
x2 + 3x + 3 = 7Subtract 7 from each side:x2 + 3x - 4 = 0Factor the left side:(x + 4) (x - 1) = 0x = - 4x = +1
(x - 1)(x - 2)
x2 + x + 6
- If 3X - 1 = 11, what is the value of X^2 + X?