Question

If α and β are the zeroes of polynomial p(x) = 3x² + 2x + 1, whose zeroes are (1 - α/1 + α) and  (1 - β/1 + β)

Answer 1

Since α and β are the zeroes of polynomial p(x) = 3x² + 2x + 1

Hence, 

\(\alpha + \beta = - \frac 23\)

and \(\alpha \beta= \frac 13\)

Now for the new polynomial,

Sum of the zeroes = \(\frac{1 - \alpha}{1 + \alpha} + \frac{1 - \beta }{1 + \beta }\) 

\(= \frac{(1 - \alpha + \beta - \alpha\beta)+(1 + \alpha - \beta - \alpha \beta)}{(1 + \alpha )(1 + \beta)}\)

\(= \frac{2 - 2\alpha \beta}{1 + \alpha + \beta +\alpha \beta}\)

\(= \cfrac {2 - \frac 23}{1 - \frac 23 + \frac 13}\)

\(= \cfrac{\frac43}{\frac 23}\)

\(= 2\)

Product of zeroes = \(\left[\frac{1 - \alpha}{1 + \alpha}\right] \left[\frac{1 - \beta }{1 + \beta }\right]\)

\(= \frac{(1 - \alpha)(1 - \beta)}{(1 + \alpha)(1 +\beta)}\)

\(= \frac{1 - \alpha - \beta + \alpha \beta}{1 + \alpha + \beta + \alpha\beta}\)

\(= \frac{1 - (\alpha + \beta)+ \alpha\beta}{1 + (\alpha + \beta) + \alpha \beta}\)

\(= \cfrac{1 + \frac 23 + \frac 13}{1 - \frac 23 + \frac 13}\)

\(= \cfrac{\frac 63}{\frac 23}\)

\(= 3\)

Hence,

Required polynomial = x² - (Sum of zeroes)x + Product of zeroes

= x² - 2x + 5

Answer 2

Since α and β are the zeroes of polynomial p(x) = 3x² + 2x + 1

Comments

Thank you so much @Ria
For answering this question