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When x3 plus 3x2-2x plus 7 is divided by x plus 1 what is the remainder?
(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)
What are the quotient and remainder when x cubed plus 3x squared plus 6x plus 1 is divided by x plus 1?
x3+3x2+6x+1 divided by x+1 Quotient: x2+2x+4 Remaider: -3
What is X3 plus X2 plus X plus 1 divided by X-1?
(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)
What times x2-x3?
-1
X3 plus x2 minus 6x plus 4?
x3 + x2 - 6x + 4 = (x - 1)(x2 + 2x - 4)
