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X^2-9x-4
They are terms that have the same variables like x, 2x and 3x or x2, 2x2 and 3x2
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)
7x^4+2x^3-x^2-x+1 (Apex ;D)If for some reason this isn't the answer then I am very sorry...
2X - 4 = -10 2X = -6 X = -3
3-2x=7 so -2x=4 and x=-2
2x2 + x2 = 3x2
(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)
They are terms that have the same variables like x, 2x and 3x or x2, 2x2 and 3x2
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)
NOT CALCULUS. Use long division. 2x2 + x - 1(4x^4) 2x^2 goes into 4x^4 2x^2 times. The remainder will then be (2x^2*(x-1)). This result(2x^3 - 2x^2) is what need to be subtracted from 4x^4 to make it exactly divisible
2x2 x 3x2 = 6 x4 (2x)2 x (3x)2 = 36 x4
They are terms that have the same variables like x, 2x and 3x or x2, 2x2 and 3x2
2x2 + (5x-4x)2 + (x-x)2 = 2x2 + x2 + 02 = 2x2 + x2 = 3x2
x2 - (2x)2 = x2 - 4x2 = -3x2 x2 - 2x2 = -x2
What is the result of adding -3x²-5x +1 and 8x²-2x-9?
2x2- 2x = 2x*(x - 1)
The question is ambiguous and two possible answers are given below: 2x2 - x2 = x2 or (2x)2 - x2 = 4x2 - x2 = 3x2