If the zeroes of the quadratic polynomial 𝑥² + (𝑎 + 1)𝑥 + 𝑏 are 2 and − 3, then find the value of 𝑎 and 𝑏.

We have been given the zeroes of the quadratic polynomial  𝑥² + (𝑎 + 1)𝑥 + 𝑏.

Let the polynomial be 𝑓(𝑥)

⇒𝑓(𝑥)= 𝑥² + (𝑎 + 1)𝑥 + b

The zeroes of a polynomial satisfy it when plugged in the equation. Since 2 and − 3 are the zeroes of the quadratic polynomial,

⇒ 𝑓(2) = 0 and, 𝑓(− 3) = 0

$\begin{array}{r}\text{ Now, }f\left(2\right)=0\\ \Rightarrow 2^2+(a+1)\times 2+b=0\\ \Rightarrow 6+2a+b=0\\ \Rightarrow 2a+b=-6\end{array}$

$\begin{array}{r}\Rightarrow (-3{)}^2+(a+1)\times -3+b=0\\ \Rightarrow 6-3a+b=0\\ \Rightarrow b-3a=-6\\ \Rightarrow 3a-b=6\end{array}$

Adding (𝑖) and (𝑖𝑖), 

we get 5𝑎 = 0 

⇒ 𝑎 = 0

Putting 𝑎 = 0 in equation (𝑖), 

we get 2 × 0 + 𝑏 =− 6

 ⇒ 𝑏 =− 6 

Hence, the value of 𝑎 is 0 and 𝑏 is − 6.