# Polynomials

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This question is one of the tougher examples of remainder theorem but it can be solved nonetheless. Check it out!!

## Polynomial Remainder Theorem

Solve the inequality x3 – 5x2 + 8x – 4 > 0.
1. (2, ${\infty}$)
2. (1, 2) ${\cup}$ (2, ${\infty}$)
3. (- ${\infty}$, 1) ${\cup}$ (2, ${\infty}$)
4. (- ${\infty}$, 1)

Choice B. (1, 2) ${\cup}$ (2, ${\infty}$)

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## Detailed Solution

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

Correct Answer: (1, 2) ${\cup}$ (2, ${\infty}$).

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