Question

.If the zeroes of the quadratic polynomial f(x) = x2 + (a + 1)x + b are 2, -3 the find the value of a
and b.​

Answer 1

$0=2^2+(a+1)\cdot 2+b\\0=(-3)^2+(a+1)\cdot(-3)+b\\\\0=4+2a+2+b\\0=9-3a-3+b\\\\b=-2a-6\\b=3a-6\\\\-2a-6=3a-6\\5a=0\\a=0\\\\b=3\cdot0-6=-6\\\\\boxed{a=0,b=-6}$

Answer 2

$\Large{\textbf{\underline{\underline{According\;to\;the\;Question}}}}$

Here,

P(x) = x² + (a + 1)x + b

Given,

P(2) = 0 & P(-3) = 0

Here,

Using value of P(2),

(2)² + (a + 1)(2) + b = 0

4 + 2a + 2 + b = 0

6 + 2a + b = 0

2a + b = -6 ..................... (1)

Now

Using value of P(-3),

(-3)² + (a + 1)(-3) + b = 0

9 - 3a - 3 + b = 0

6 - 3a + b = 0

-3a + b = -6 ................... (2)

Now,

Subtract (1) & (2),

2a + b - (-3a + b) = - 6 - (-6)

2a + b + 3a - b = -6 + 6

5a = 0

a = 0

Now,

Substitute the value of a in (1),

2a + b = -6

2(0) + b = -6

b = -6