Question

if alpha and beta are zeroes of polynomial 3x^2-2x-8 find the polynomial whose zero are 1/alpha and 1/beta​

Answer 1

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Answer 2

The polynomial is :-

P(x) 3x² - 2x - 8

Factorising the polynomial :-

$= 3 {x}^{2} - 2x - 8 \\ \\ \\ = 3 {x}^{2} - 6x + 4x - 8 \\ \\ \\ = 3x(x - 2) + 4(x - 2) \\ \\ \\ = (x - 2)(3x + 4)$

This implies that the zeroes of the given polynomial is :-

$x - 2 = 0 \\ x = 2 \\ \\ \\ 3x + 4 = 0 \\ \\ x = \frac{ - 4}{3}$

$\rule{200}{2}$

Now,

We want a polynomial with the zeroes as 1/alpha and 1/beta.

Thus,

Zeroes of the new polynomial is :-

$\frac{1}{2} \\ \\ \\ \frac{ - 3}{4}$

Now,

Sum of zeroes :-

$\frac{1}{2} - \frac{3}{4} \\ \\ \\ = \frac{2 - 3}{4} \\ \\ \\ = \frac{ - 1}{4}$

Product of zeroes :-

$( \frac{1}{2})( \frac{ - 3}{4}) \\ \\ \\ = \frac{ - 3}{8}$

Now,

The new polynomial can be :-

$p(x) = k( {x}^{2} + \frac{1}{4}x - \frac{3}{8}) \\ \\ \\ \\ \\ = 8 {x}^{2} + 2x - 3$

When k is equal to 8.

There can be many number of polynomial with the given zeroes.

Hence,

Solved!