Question

P(x)= x^2+5x+6 find the zeros of the polynomial and verify the relationship between zeros and its coefficient

Answer 1

x²+5x+6

=x²+3x+2x+6

=x{x+3}+2{x+3}

={x+2}{x+3}

So, x=-2,-3

α=-2,β=-3

α+β=-coefficient of x/coefficient of x²

{-2-3}=-{5/1}

-5=-5

αβ=constant /coefficient of x²

-2×-3=6

6=6

pe it helps u ...

Answer 2

Given :-

p ( x ) = x² + 5x + 6

Required to find :-

• Zeroes of the polynomial ?

• Verify the relationship between the zeroes and it's coefficients

Solution :-

Given data :-

p ( x ) = x² + 5x + 6

we need to find the zeroes of the polynomial and verify the relationship between zeroes and it's coefficients .

So,

Let's Factorise the given quadratic polynomial ;

p ( x ) = x² + 5x + 6

x² + 5x + 6

x² + 3x + 2x + 6

x ( x + 3 ) + 2 ( x + 3 )

( x + 3 ) ( x + 2 )

This implies ;

x + 3 = 0

x = - 3

α = - 3

Similarly,

x + 2 = 0

x = - 2

β = - 2

Hence,

Zeroes of the polynomial are - 3 and - 2

Now,

Let's verify the relationship between the zeroes and it's coefficients

The relation between the sum of the zeroes and the coefficients is ;

$\boxed{ \tt {\pink{ \alpha + \beta = \dfrac{ - \: coefficient \: of \: x }{coefficient \: of \: {x}^{2} } }}}$

α + β =

=> - 3 + ( - 2 )

=> - 3 - 2

=> - 5

But,

$\sf{ \dfrac{ - \: coefficient \: of \: x}{coefficient \: of \: {x}^{2} }}$

=> - ( 5 ) / 1

=> - 5/1

=> - 5

Hence,

$\boxed{ \tt { \alpha + \beta = \dfrac{ - \: coefficient \: of \: x }{coefficient \: of \: {x}^{2} } }}$

Similarly,

The relation between the product of the zeroes and the coefficients is ;

$\boxed{ \tt{ \green{ \alpha \beta = \frac{constant \: term}{coefficient \: of \: {x}^{2} } }}}$

α β =

=> - 3 x - 2

=> 6

But,

$\rm{ \dfrac{constant \: term}{coefficient \: of \: {x}^{2} }}$

=> 6/1

=> 6

Hence,

$\boxed{ \tt{ \alpha \beta = \frac{constant \: term}{coefficient \: of \: {x}^{2} } }}$

Therefore,

The relation between the zeroes and it's coefficients had been verified .