For the following, find a quadratic polynomial whose sum and product respectively of the zeros are as given. Also, find the zeros of these polynomials by factorization.
$-\frac{8}{3},\ \frac{4}{3}$

Given:

Sum of the zeros of a polynomial$=-\frac{8}{3}$.

Product of the zeros of the polynomial$=\frac{4}{3}$.

To do:

Here, we have to find the quadratic polynomial whose sum and product of the zeros are as given.

Solution:

A quadratic polynomial formed for the given sum and product of zeros is given by:

$f(x) = x^2 -(sum of zeros) x + (product of zeros)$

Therefore,

The required polynomial f(x) is,

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

$=x^2 + \frac{8}{3}x + \frac{4}{3}$

To find the zeros of f(x), we put $f(x) = 0$.

This implies, 

$x^2 + \frac{8}{3}x + \frac{4}{3} = 0$

Multiplying by 3 on both sides, we get, 

$3(x^2) + 3(\frac{8}{3})x + 3(\frac{4}{3})= 0$

$3x^2+8x+4=0$

$3x^2 + 6x + 2x + 4 = 0$

$3x(x + 2) + 2(x + 2) = 0$

$(x + 2) (3x + 2) = 0$

$(x + 2) = 0$ and $(3x + 2) = 0$

$x=-2$ and $x=-\frac{2}{3}$

Therefore, the two zeros of the quadratic polynomial are $-2$ and $-\frac{2}{3}$.

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Updated on: 10-Oct-2022

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