x3+3x2+6x+1 divided by x+1 Quotient: x2+2x+4 Remaider: -3
x3+3x2+3x+2 divided by x+2 equals x2+x+1
x3+3x2-22x-90 divided by x-5 equals x2+8x+18 For the answer to be correct then the quotient multiplied by the divisor must be equal to the dividend. Hence:- (x-5)*(x2+8x+18) Multiplying out the brackets term by term: x3-5x2+8x2-40x+18x-90 Collecting like terms in descending order: x3+3x2-22x-90 So the answer is correct.
(x3 + x2 + x + 1)/(x -1) (using the long division)x2(x - 1) = x3 - x2x3 + x2 + x + 1 - (x3 - x2) = 2x2 + x + 12x(x - 1) = 2x2 - 2x2x2 + x + 1 - (2x2 - 2x) = 3x + 13(x - 1) = 3x - 33x + 1 - (3x - 3) = 4 (the remainder)(x3 + x2 + x + 1)/(x -1) = x2 + 2x + 3 + 4/(x -1)(1x3 + 1x2 + 1x + 1)/(x -1) (using the synthetic division)(the constant of the divisor) 1] 1 1 1 1 (the coefficients of the dividend)The coefficients of the quotient:11 + 1*1 = 21 + 2*1 = 3Since the degree of the first term of the quotient is one less than the degree of the first term of the dividend, the quotient is x2 + 2x + 3.The remainder1 + 3*1 = 4(x3 + x2 + x + 1)/(x -1) = x2 + 2x + 3 + 4/(x -1)
If you mean: y =(lnx)3 then: dy/dx = [3(lnx)2]/x ddy/dx = [(6lnx / x) - 3(lnx)2] / x2 If you mean: y = ln(x3) Then: dy/dx = 3x2/x3 = 3/x = 3x-1 ddy/dx = -3x-2 = -3/x2
(x3 - 3x2 + 4x - 7) - (2x3 + x2 - 3x - 5)
(x3 + 4x2 - 3x - 12)/(x2 - 3) = x + 4(multiply x2 - 3 by x, and subtract the product from the dividend)1. x(x2 - 3) = x3 - 3x = x3 + 0x2 - 3x2. (x3 + 4x2 - 3x - 12) - (x3 + 0x2 - 3x) = x3 + 4x2 - 3x - 12 - x3 + 3x = 4x2 - 12(multiply x2 - 3 by 4, and subtract the product from 4x2 - 12)1. 4x(x - 3) = 4x2 - 12 = 4x2 - 122. (4x2 - 12) - (4x2 - 12) = 4x2 - 12 - 4x2 + 12 = 0(remainder)
x3 + ax + 3a + 3x2 = x (x2 + a) + 3 (a + x2) = x (x2 + a) + 3 (x2 + a) = (x2 + a)(x + 3) Checking the work: x3 + ax + 3x2 + 3a or x3 + 3x2 + 3a + ax = x2 (x + 3) + a (3 + x) = x2 (x + 3) + a (x + 3) = (x + 3)(x2 + a)
x3+3x2+6x+1 divided by x+1 Quotient: x2+2x+4 Remaider: -3
3 - 3x + x2 - x3 = (1 - x)(x2 + 3)
x3 + x2 - 3x - 3 x(x2 + x - 3) - 3
x3 - 3x2 + x - 3 = (x2 +1)( x - 3)
x3 - 3x2 + x - 3 = (x - 3)(x2 + 1)
x3+3x2+3x+2 divided by x+2 equals x2+x+1
Suppose the middle one is 3x, then the other two are 3x-3 and 3x+3. Their product is (3x-3)*3x*(3x+3) = 3x*(3x-3)*(3x+3) = 27x(x-1)*(x+1) = 27x*(x2-1) So 27x*(x2-1) = 648 x*(x2-1) = 648/27 = 24 ie x3 - x - 24 = 0 or x3 - 3x2 + 3x2 - 9x + 8x - 24 = 0 ie (x - 3)(x2 + 3x + 8) = 0 The only integer solution is x = 3 So the three numbers are 3*(3-1), 3*3 and 3*(3+1) = 6, 9 and 12.
x3+3x2-22x-90 divided by x-5 equals x2+8x+18 For the answer to be correct then the quotient multiplied by the divisor must be equal to the dividend. Hence:- (x-5)*(x2+8x+18) Multiplying out the brackets term by term: x3-5x2+8x2-40x+18x-90 Collecting like terms in descending order: x3+3x2-22x-90 So the answer is correct.
x3 + 3x2 - 4x -12 x2(x + 3) - 4(x + 3) (x2 - 4)(x+3)