Question

For each of the following, find a quadratic polynomial whose sum and product respectively of the zeroes are as given. Also, find the zeroes of these polynomials by factorization: -8/4, 4/3

Answer 1

In the above Question , the following information is given -

Instructions -

For each of the following, find a quadratic polynomial whose sum and product respectively of the zeroes are as given.

Also, find the zeroes of these polynomials by factorization .

Now ,

The given Zeroes of the polynomial are provided to be -

Zero 1 -

Alpha = ( - 8 / 4 )

Beta = ( 4 / 3 )

Sum Of Zeroes

=> Alpha + Beta

=> ( -8 / 4 ) + ( 4 / 3 )

=> -2 + ( 4 / 3 )

=> ( 4 / 3 ) - 2

=> ( -2 / 3 )

Product Of Zeroes -

=> Alpha × Beta

=> [ -8 / 4 ] × [ 4 / 3 ]

=> [ -8 / 3 ] × [ 4 / 4 ]

=> [ -8 / 3 ]

Now , a quadratic polynomial can be written as -

x² - [ Alpha + Beta ]x + [ Alpha × Beta ] = 0

=> x² - [ -2 / 3 ] x + [ -8 / 3 ] = 0

=> x² + [ 2 x / 3 ] - [ 8 / 3 ] = 0

=> 3 x² + 2x - 8 = 0

Thus , the required polynomial is 3x² - 2x - 8 .

Factorisation -

3x² + 2x - 8

=> 3x² + 6x - 4x - 8

=> 3x ( x + 2 ) - 4 ( x + 2 )

=> ( 3x - 4 )( x + 2 )

Thus,

Required Zeroes - ( 3 / 4 ) , -2

Hence Verified

_________________

Answer 2

$\huge\sf\pink{Answer}$

☞ $\sf{ P(x) = 3x^2 + 8x + 4}$

☞ Zeros = (-2) or (-2/3)

$\rule{110}1$

$\huge\sf\blue{Given}$

✭ $\sf Sum \: of \: zeros = \dfrac{-8}{4}$

✭ $\sf Product\: zeros = \dfrac{4}{3}$

$\rule{110}1$

$\huge\sf\gray{To \:Find}$

☆ The polynomial and it's zeros

$\rule{110}1$

$\huge\sf\purple{Steps}$

We know that,

P(x) = x² - (α+β)x + αβ

Substituting the given values,

➝ $\sf P(x) = x^2 - (\dfrac{-8}{4})x + \dfrac{4}{3}$

➝ $\sf P(x) = x^2 + \dfrac{8}{4}x + \dfrac{4}{3}$

➝ $\sf\color{red}{ P(x) = 3x^2 + 8x + 4}$

Finding the zeros of the polynomial,

$\sf\dashrightarrow 3x^2+8x+4$

$\sf\dashrightarrow 3x^2+ 6x+2x+4$

$\sf\dashrightarrow 3x(x+2) + 2(x+2)$

$\sf\orange{\dashrightarrow (3x+2)(x+2)}$

Equating with 0,

$\leadsto\sf 3x +2 = 0$

$\sf\color{lime}{ \leadsto x = \dfrac{-2}{3}}$

Or

$\sf\longmapsto x+2 = 0$

$\sf\color{aqua}{\longmapsto x = -2}$

$\rule{170}3$