Question

If \(\alpha, \beta,\) and \(\gamma\) are the zeros of the polynomial \(6x^3 - 29x^2 + 46x + 24\), then the polynomial whose zeros are \(\frac1\alpha,\frac1\beta,\frac1\gamma\) is:

• A. \(29x^3+51x^2-34x+1\)
• B. \(24x^3+46x^2-29x+6\)
• C. \(21x^3+43x^2-26x+3\)
• D. \(10x^3+32x^2-37x-8\)

Hint:

For reciprocal roots, reverse the coefficients of the polynomial and retain their signs. This shortcut works very efficiently in examinations.

Answer 1

The Correct Option is B

Concept: If \[ P(x)=a_0x^n+a_1x^{n-1}+\cdots+a_n \] has roots \(\alpha,\beta,\gamma\), then the polynomial whose roots are \[ \frac1\alpha,\frac1\beta,\frac1\gamma \] is obtained by replacing \(x\) by \(\frac1x\) and multiplying by \(x^n\).

Step 1: Write the given polynomial. \[ P(x)=6x^3-29x^2+46x+24 \]

Step 2: Substitute \(x=\frac1t\). \[ P\!\left(\frac1t\right) = 6\left(\frac1{t^3}\right) - 29\left(\frac1{t^2}\right) + 46\left(\frac1t\right) + 24 \] Multiplying by \(t^3\), \[ 6-29t+46t^2+24t^3 \]

Step 3: Arrange in descending powers. \[ 24t^3+46t^2-29t+6 \] Replacing \(t\) by \(x\), \[ \boxed{24x^3+46x^2-29x+6} \]