Question

If \(\alpha, \beta, \gamma, \delta, \epsilon\) are the roots of the equation \(x^5 + x^4 - 13x^3 - 13x^2 + 36x + 36 = 0\) and \(\alpha<\beta<\gamma<\delta<\epsilon\) then \( \frac{\epsilon}{\alpha} + \frac{\delta}{\beta} + \frac{1}{\gamma} = \)

• A. 0
• B. 1
• C. -1
• D. -3

Hint:

When factoring high-degree polynomials, always look for simple patterns like factoring by grouping. If the powers of the variable decrease in a regular way (e.g., x^4, x^2, constant), you can often make a substitution (like y=x^2) to reduce it to a simpler polynomial.

Answer 1

The Correct Option is D

Let the given polynomial be \(P(x) = x^5 + x^4 - 13x^3 - 13x^2 + 36x + 36\).
We can factor this polynomial by grouping terms.
\(P(x) = x^4(x+1) - 13x^2(x+1) + 36(x+1)\).
We can factor out the common term \((x+1)\).
\(P(x) = (x+1)(x^4 - 13x^2 + 36)\).
So, one root is \(x = -1\).
Now we need to solve the remaining quartic equation: \(x^4 - 13x^2 + 36 = 0\).
This is a quadratic equation in terms of \(x^2\). Let \(y = x^2\).
The equation becomes \(y^2 - 13y + 36 = 0\).
Factoring the quadratic: \((y-9)(y-4) = 0\).
So, the possible values for \(y\) are \(y=9\) or \(y=4\).
Substituting back \(y = x^2\), we have \(x^2=9\) or \(x^2=4\).
From \(x^2=9\), we get roots \(x=3\) and \(x=-3\).
From \(x^2=4\), we get roots \(x=2\) and \(x=-2\).
The five roots of the polynomial are \(-3, -2, -1, 2, 3\).
The problem states that the roots are ordered: \(\alpha<\beta<\gamma<\delta<\epsilon\).
So, we have \(\alpha = -3\), \(\beta = -2\), \(\gamma = -1\), \(\delta = 2\), and \(\epsilon = 3\).
Now, we evaluate the expression \( \frac{\epsilon}{\alpha} + \frac{\delta}{\beta} + \frac{1}{\gamma} \).
\( \frac{3}{-3} + \frac{2}{-2} + \frac{1}{-1} = -1 + (-1) + (-1) = -3 \).