3x2 + 2x + 3 + x2 + x + 1 = 4x 2+ 3x + 4
x3 - x2 - 2x +2 you can divide this into two parts for now x3 - x2 and -2x +2 factor out everything you can from both x2(x-1) and -2(x+1) since the insides of the parentheses are the same, we can put them together, then multiply it by the sum of the two parts outside the parentheses. (x-1)(x2-2)
x2 + 2x - 3 = (x + 3)(x - 1)
3x3 + 6x2 + x + 2 3x2 (x + 2) + (x + 2) (x + 2) (3x2 + 1) OR 3x3 + x + 6x2 + 2x (3x2 + 1) + 2 (3x2 + 1) (3x2 + 1) (x + 2) Check: (x + 2) (3x2 + 1) = 3x3 + x + 6x2 + 2 = 3x3 + 6x2 + x + 2 x3 - 3x2 - 4x + 12 x2 (x - 3) - 4 (x - 3) (x - 3) (x2 - 4) (x - 3) (x + 2) (x - 2) Check: (x - 3) (x + 2) (x - 2) = (x - 3) (x2 - 4) = x3 - 4x - 3x2 + 12= x3 - 3x2 - 4x + 12=== === x5 - x4 + 8x3 - 8x2 + 16x - 16 x4 (x - 1) + 8x2 (x - 1) + 16 (x - 1) (x - 1) (x4 + 8x2 + 16) (x - 1) (x2 + 4)(x2 + 4) (x - 1) (x2 + 4)2 Real Solution (x - 1) (x + 2i) (x - 2i) (x + 2i) (x - 2i) (x - 1) (x + 2i)2 (x - 2i)2 Check: (x - 1) (x + 2i)2 (x - 2i)2 = (x - 1) (x + 2i) (x - 2i) (x + 2i) (x - 2i) = (x - 1) (x2 + 2xi - 2xi - 4i2) (x2 + 2xi - 2xi - 4i2) = (x - 1) (x2 - 4(-1)) (x2 - 4(-1)) = (x - 1) (x2 + 4) (x2 + 4) = (x - 1) (x4 + 4x2 + 4x2 +16) = (x - 1) (x4 + 8x2 + 16) = x5 + 8x3 +16x - x4 - 8x2 - 16 = x5 - x4 + 8x3 - 8x2 + 16x - 16
x3 - 3x2 + x - 3 = (x2 +1)( x - 3)
3x2 + 2x + 3 + x2 + x + 1 = 4x 2+ 3x + 4
3x3 - x2 - x - 1 = 3x3 - 3x2 + 2x2 - 2x + x - 1 = 3x2(x - 1) + 2x(x - 1) + 1(x - 1) = (3x2 + 2x + 1)(x - 1) So 3x3 - x2 - x - 1 /(x - 1) = (3x2 + 2x + 1)
(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/(3x2 - 5x - 2) - [2x/ (3x + 1)][1/(x - 2)] = x2/(3x2 - 6x + x - 2) - 2x/(3x + 1)(x - 2) = x2/[3x(x - 2) + (x - 2)] - 2x/(3x + 1)(x - 2) = x2/(3x + 1)(x - 2) - 2x/(3x + 1)(x - 2) = (x2 - 2x)/(3x + 1)(x - 2) = x(x - 2)/(3x + 1)(x - 2) = x/(3x + 1)
(x3 + 3x2 - 2x + 7)/(x + 1) = x2 + 2x - 4 + 11/(x + 1)(multiply x + 1 by x2, and subtract the product from the dividend)1. x2(x + 1) = x3 + x22. (x3 + 3x2 - 2x + 7) - (x3 + x2) = x3 + 3x2 - 2x + 7 - x3 - x2 = 2x2 - 2x + 7(multiply x + 1 by 2x, and subtract the product from 2x2 - 2x + 7)1. 2x(x + 1) = 2x2 + 2x2. (2x2 - 2x + 7) - (2x2 + 2x) = 2x2 - 2x + 7 - 2x2 - 2x = -4x + 7(multiply x + 1 by -4, and subtract the product from -4x + 7)1. -4(x + 1) = -4x - 42. -4x + 7 - (-4x - 4) = -4x + 7 + 4x + 4 = 11(remainder)
x2+8x = -3x2+5 4x2+8x-5 = 0 (2x+5)(2x-1) = 0 x = -5/2 or x = 1/2
3x2 + 2x + 2 = 0 Can not be factored. You can however solve it for x: 3x2 + 2x = -2 x2 + 2x/3 = -2/3 x2 + 2x/3 + 1/9 = -2/3 + 1/9 (x + 1/9)2 = -5/9 x + 1/9 = ±√(-5/9) x = -1/9 ± i√(5/9) x = -1/9 ± i√5 / 3 x = -1/9 ± 3i√5 / 9 x = (-1 ± 3i√5) / 9
4x2 -x -1 :D
x2 + x2 + x2 = (1 + 1 + 1)x2 = 3x2
x^5+2x^4+4x^2+2x-3
What do you want to convert it to? x2 + y2 = 2x If you want to solve for y: x2 + y2 = 2x ∴ y2 = 2x - x2 ∴ y = (2x - x2)1/2 If you want to solve for x: x2 + y2 = 2x ∴ x2 - 2x = -y2 ∴ x2 - 2x + 1 = 1 - y2 ∴ (x - 1)2 = 1 - y2 ∴ x - 1 = ±(1 - y2)1/2 ∴ x = 1 ± (1 - y2)1/2
x3 - x2 - 2x +2 you can divide this into two parts for now x3 - x2 and -2x +2 factor out everything you can from both x2(x-1) and -2(x+1) since the insides of the parentheses are the same, we can put them together, then multiply it by the sum of the two parts outside the parentheses. (x-1)(x2-2)