The question is about Polynomial Remainder Theorem. We need to find out the range which is also a solution to the given inequality. This question is one of the tougher examples of remainder theorem but it can be solved nonetheless. Check it out! Polynomials is a simple topic which involves a lot of basic ideas. Make sure that you a get of hold of them by solving these questions.

Question 2: Solve the inequality x^{3} – 5x^{2} + 8x – 4 > 0.

- (2,\\infty \\))
- (1, 2) \\cup\\) (2, \\infty\\))
- (-\\infty \\), 1) \\cup\\) (2, \\infty \\))
- (-\\infty \\), 1)

(1, 2) \\cup\\) (2, \\infty\\))

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Let a, b, c be the roots of this cubic equation

a + b + c = 5

ab + bc + ca = 8

abc = 4

This happens when a = 1, b = 2 and c = 2 {This is another approach to solving cubic equations.}

The other approach is to use polynomial remainder theorem.

If you notice, sum of the coefficients = 0

=> P(1) = 0

=> (x – 1) is a factor of the equation. Once we find one factor, we can find the other two by dividing the polynomial by (x–1) and then factorising the resulting quadratic equation.

(x – 1) (x – 2) (x – 2) > 0

Let us call the product (x – 1)(x – 2)(x – 2) a black box.

If x is less than 1, the black box is a –ve number.

If x is between 1 and 2, the black box is a +ve number.

If x is greater than 2, the black box is a +ve number.

Since we are searching for the regions where black box is a +ve number, the solution is as follows:

1 < x < 2 OR x > 2

The question is **"Solve the inequality x ^{3} – 5x^{2} + 8x – 4 > 0."**

Choice B is the correct answer.

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