Question

if the zeros of polynomial x^2+(a+1)x+b are 2 and -5 then find the value of and b​

Answer 1

Answer:

we know that 2and-5 are zeroes of polynomial.

then,

we have learnt in ch-polynomials relation between zeroes of polynomial.

alpha (A)=2. and beta(B)= -5.

A+B= -(a+1)/1. Ist relation. putting values

-3= -a-1

hence a=2 ans

IInd relation=A multiply by B=b/1. putting values

-10=b ans

Answer 2

Step-by-step explanation:

Given -

Zeroes are 2 and -5 of polynomial

p(x) = x² + (a + 1)x + b

To Find -

Value of a and b

Now,

p(2) = (2)² + (a + 1)2 + b

= 4 + 2a + 2 + b = 0

= 2a + b = - 6 ........ (a)

And

p(-5) = (-5)² + (a + 1)-5 + b

= 25 - 5a - 5 + b = 0

= - 5a + b = - 20 ......... (b)

Subtracting (a) and (b)

2a + b = - 6

-5a + b = - 20

(+) (-) (+)

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

7a = 14

a = 14/7

• a = 2

Substituting the value of a on 2a + b = - 6

= 2(2) + b = - 6

= 4 + b = - 6

= b = - 6 - 4

• = b = - 10

Hence,

The value of a and b is

a = 2 and b = - 10

Verification -

Substituting the value of a and b on

x² + (a + 1)x + b

The polynomial is x² + (2 + 1)x + (-10)

• x² + 3x - 10

= x² + 3x - 10

= x² - 2x + 5x - 10

= x(x - 2) + 5(x - 2)

= (x + 5)(x - 2)

Zeroes are -

x + 5 = 0 and x - 2 = 0

x = -5 and x = 2

Zeroes comes same as given, it shows that our answer is absolutely correct.

Hence,

verified..