Question

If zeroes of a polynomial are 4 and-3 then find the polynomial

Answer 1

Answer :-

${x}^{2} - x + 12$

Given :-

Zeroes of polynomial :- 4 , -3

To find :-

The required polynomial

Solution :-

Let $\alpha$ and $\beta$ be the zeroes of polynomial.

then ,

→sum of zeroes ,

$\alpha + \beta = 4 - 3$

$\alpha + \beta = 1$

→Product of zeroes ,

$\alpha \beta = 4 \times - 3$

$\alpha \beta = - 12$

Required polynomial →

$= > {x}^{2} - ( \alpha + \beta )x + \alpha \beta$

$= > {x}^{2} - (1) x+ ( - 12)$

$= > {x}^{2} - x + 12$

hence , required polynomial will be${x}^{2} - x + 12$

Answer 2

Answer:

$\implies\ x^2 - x +12$

Step-by-step explanation:

Given Problem:

If zeroes of a polynomial are 4 and-3 then find the polynomial.

Solution:

To Find:

The polynomial.

---------------

Method:

According to your question,

Zeroes of polynomial : 4 , -3

We know that,

Sum of zeroes = Alpha +Beta

Product of zeroes = Alpha.Beta

Let $\alpha$ and $\beta$ be the zeroes of polynomial.

Now,

Sum of Zeroes,

$\implies\alpha +\beta = 4-3$

$\implies\alpha + \beta = 1$

Now,

Product of zeroes,

$\implies\alpha \beta = 4 \times\ -3$

$\implies\alpha\beta = -12$

Now Polynomial,

$\implies\ x^2 - (\alpha + \beta )\ x+ \alpha\beta$

$\implies\ x^2 - (1)\x+(-12)$

$\implies\ x^2 - x+12..............(Answer)$