What is the answer for question 8. b) ?

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You need to evaluate the polynomial inequality, such that:

`x^3 - 5x^2 + 2x >= -8`

Taking all terms to one side, yields:

`x^3 - 5x^2 + 2x + 8 >= 0`

You need to attach the equation and solve it for `x` , such that:

`x^3 - 5x^2 + 2x + 8 = 0`

Testing if `x = 2` is a solution to equation yields:

`8 - 20 + 4 + 8 = 0 => 0 = 0 => x = 2` is a solution to equation `x^3 - 5x^2 + 2x + 8 = 0.`

You may convert the polynomial in its factored form, such that:

`x^3 - 5x^2 + 2x + 8 = (x - 2)(ax^2 + bx + c)`

Performing the multiplication to the right, yields:

`x^3 - 5x^2 + 2x + 8 = ax^3 + bx^2 + cx - 2ax^2 - 2bx - 2c`

`x^3 - 5x^2 + 2x + 8 = ax^3 + x^2(b - 2a) + x(c - 2b) - 2c`

Equating the coefficients of like powers yields:

`a = 1`

`b - 2a = -5 => b - 2 = -5 => b = -3`

`c - 2b = 2 => c + 6 = 2 => c = -4`

`-2c = 8 => c =- 4`

Hence, using the method of undetermined coefficients, yields:

`ax^2 + bx + c = x^2 - 3x - 4`

`x^3 - 5x^2 + 2x + 8 = (x - 2)(x^2 - 3x - 4)`

You need to test if the quadratic `x^2 - 3x - 4 = 0` can be converted in its factored form ` x^2 - 3x - 4 = (x - x_1)(x - x_2` ), using the quadratic formula, such that:

`x_(1,2) = (3+-sqrt(9 + 16))/2 => x_(1,2) = (3+-5)/2`

`x_1 = 4 ; x_2 = -1`

`x^2 - 3x - 4 = (x - 4)(x + 1)`

`x^3 - 5x^2 + 2x + 8 = (x + 1)(x - 2)(x - 4)`

You need to discuss the sign of expression `x^3 - 5x^2 + 2x + 8 ` over the intervals `(-oo,-1],(-1,2],(2,4],(4,oo)` , such that:

If `x in (-oo,-1] => x^3 - 5x^2 + 2x + 8 <= 0`

If `x in (-1,2] => x^3 - 5x^2 + 2x + 8 >= 0`

If `x in (2,4] => x^3 - 5x^2 + 2x + 8 <= 0`

If `x in (4,oo) => x^3 - 5x^2 + 2x + 8 >= 0`

**Hence, solving the given polynomial inequality yields that it holds for `x in [-1,2]U[4,oo)` .**

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