Best Answer

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) = 0

x(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) = 0

x + 1 = 0 or 2x2 - 2x - 3 = 0

x = -1 or

x = [-2 +/- square root of ((-2)2 - 4(2)(-3))]/2(2)

x = [-2 +/- square root of (4 + 24)]/4

x = -1/2 +/- (square root of 28)/4

x = -1/2 +/- 1/2(square root of 7)

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Q: X4 plus 3x3 - x2 - 9x - 6 equals 0?

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Answer this question…A. x4 + 2x3 + 9x2 + 4 B. x4 + 4x3 + 9x2 + 4 C. x4 + 2x3 + 9x2 + 4x + 4 D. x4 + 2x3 + 9x2 - 4x + 4

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)

(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 + 40Râˆ´ x4 - 2x3 + 2x2 + x + 4 = (x2 + x + 1)(x2 - 3x + 4)

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

No.

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Answer this question…A. x4 + 2x3 + 9x2 + 4 B. x4 + 4x3 + 9x2 + 4 C. x4 + 2x3 + 9x2 + 4x + 4 D. x4 + 2x3 + 9x2 - 4x + 4

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)

(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 + 40Râˆ´ x4 - 2x3 + 2x2 + x + 4 = (x2 + x + 1)(x2 - 3x + 4)

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

Not necessarily.

- 3X3 + X2 + 17 + 6X4Three. I will put the coefficients in parentheses.(- 3 )X3 + (1)X2 + 17 + (6)X4=========================( remember, 1 is always an implied coefficient )

No.

x4 + 2x3 - 9x2 + 18x = x(x3 + 2x2 - 9x + 18) which I do not think can be factorised further.

Given: x4 + 3x3 - 10x2 = 0 You can start by dividing both sides of the equation by x2: x2 + 3x - 10 = 0 Then factor out the left side of the equation into simpler terms: (x + 5)(x - 2) = 0 For the result to be 0, either of the terms in brackets on the left must equal 0, so that means there are two answers: x = -5, x = 2

4X + 2x = 1. Where x = 0.166666666666666666666666666666666666666 recurring.

One of anything plus another one of them is always two of the same thing.

28 is.