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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)
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What is the anti-derivative of x2 1x?
The antiderivative of x2 + x is 1/3x3 + 1/2x2 + C.
Quotient and remainder of 3X4 plus 2X3 minus X2 minus x minus 6 divided by X2 plus 1?
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Factor x4 - 3x3 - 7x2 plus 27x -18?
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What is the derivative of x over 1 plus x squared?
The product rule states: d/dx uv = vdu/dx + udv/dxIt is not quite clear what your denominator is:"over 1 + (x squared)": d/dx (x/1+x2)d/dx(x/1+x2) = d/dx x(1 + x2)-1 = (1 + x2)-1 d/dx x + x d/dx (1 + x2)-1= (1 + x2)-1 + x -2x (1 + x2)-2= (1 + x2)-2 (1 + x2 - 2x2)= (1 - x2) / (1 + x2)2"over (1 + x) [all] squared": d/dx (x/(1+x)2)d/dx(x/(1+x)2) = d/dx x(1 + x)-2 = (1 + x)-2 d/dx x + x d/dx (1 + x)-2= (1 + x)-2 + x -2 (1 + x)-3= (1 + x)-3 (1 + x - 2x)= (1 - x)/(1 + x)3If you prefer, you can use the quotient rule: d/dx (u/v) = (v du/dx - udv/dx) / v2
What are the 4 fundamental operation in functional notation?
1) You can add/subtract functions: f(x) +- g(x) = (f +- g)(x). 2) You can multiply/divide functions: f(x) */ g(x) = (f */ g)(x). 3) You can compose functions: f(x) . g(x) = (f(g(x))) = (f . g)(x). Let f(x) = 3x + 1 and g(x) = x2 Ex 1. (f + g)(x) = x2 + 3x + 1 Ex 2. (f * g)(x) = (3x + 1) * x2 = 3x3 + x2 Ex 3. (f(g(x))) = 3(x2) + 1 (Note that you replace all the x's in the function f(x) with the whole value of g(x).
What is the anti-derivative of x2 1x?
The antiderivative of x2 + x is 1/3x3 + 1/2x2 + C.
Given 3x3 plus 4x2 plus x plus 7 is divided by x2 plus 1?
Given 3x3 + 4x2 +x + 7 is divided by x2 + 1, find the results:
X4-2x3 plus 2x2 plus x plus 4 divided by x2 plus x plus 1 solutions?
(x4 - 2x3 + 2x2 + x + 4) / (x2 + x + 1)You can work this out using long division:x2 - 3x + 4___________________________x2 + x + 1 ) x4 - 2x3 + 2x2 + x + 4x4 + x3 + x2-3x3 + x2 + x-3x3 - 3x2 - 3x4x2 + 4x + 44x2 + 4x + 40R∴ x4 - 2x3 + 2x2 + x + 4 = (x2 + x + 1)(x2 - 3x + 4)
Quotient and remainder of 3X4 plus 2X3 minus X2 minus x minus 6 divided by X2 plus 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).
What is the derivative of x to the fourth power -3x cubed plus 5x2 divided by x squared?
If you mean: f(x) = x4 - 3x3 + 5x2 / x2 then: f(x) = x4 - 3x3 + 5 ∴ f'(x) = 4x3 - 9x2 If you mean: f(x) = (x4 - 3x3 + 5x2) / x2 then: f(x) = x2 - 3x + 5 ∴ f'(x) = 2x - 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 is an example of factor by grouping?
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
What is the derivative of one third x cubed?
f(x) = 1/3x3 f'(x) = ∫ 1/3x3 dx f'(x) = 1/3∫ x3 dx f'(x) = 1/3 * 3 * x2 f'(x) = x2
What is x when X4 plus 3x3 - x2 - 9x - 6 equals 0?
x4 + 3x3 - x2 - 9x - 6 = 0 x4 + x3 + 2x3 + 2x2 - 3x2 - 3x - 6x - 6 = 0 x3(x + 1) + 2x2(x + 1) - 3x(x + 1) - 6(x + 1) = 0 (x + 1)(x3 + 2x2 - 3x - 6) = 0 (x + 1)[x2(x + 2) - 3(x + 2)] = 0 (x + 1)(x + 2)(x2 - 3) = 0 So x + 1 = 0 so that x = -1 or x + 2 = 0 so that x = -2 or x2 - 3 = 0 so that x = +/- sqrt(3)
X to the second power times -3x?
x2 * -3x = -3x3
What is the fractorization of 3x3 plus 6x2 - 24x?
3x3 + 6x2 - 24x = 3x(x2 + 2x - 8) = 3x(x2 + 4x - 2x - 8) = 3x[ x(x + 4) - 2(x + 4) ] = 3x(x - 2)(x + 4)
X4 plus 3x3 - x2 - 9x - 6 equals 0?
x4+ 3x3 - x2 - 9x - 6 = 0 (rewrite the equation, let -9x = -3x - 6x)x4+ 3x3 - x2 - 3x - 6x - 6 = 0 (group terms)(x4 - x2) + (3x3 - 3x) - (6x + 6) = 0x(x2 - 1) + 3x(x2 - 1) - 6(x + 1) = 0 (use the difference of the squares, a2 - b2 = (a - b)(a + b))x(x - 1)(x + 1) + 3x(x - 1)(x + 1) - 6(x + 1) = 0 (the common factor is x + 1)(x + 1)[x(x - 1) + 3x(x - 1) - 6] = 0(x + 1)(x2 - x + 3x2 - 3x - 6) = 0 (work with like terms)(x + 1)(4x2 - 4x - 6) = 0 (factor out 2)2(x + 1)(2x2 - 2x - 3) = 0x + 1 = 0 or 2x2 - 2x - 3 = 0x = -1 orx = [-2 +/- square root of ((-2)2 - 4(2)(-3))]/2(2)x = [-2 +/- square root of (4 + 24)]/4x = -1/2 +/- (square root of 28)/4x = -1/2 +/- 1/2(square root of 7)Check:
