If $ \left( {{x^3} + a{x^2} + bx + 6} \right) $ has $ \left( {x - 2} \right) $ as a factor and leaves remainder $ 3 $ when divided by $ \left( {x - 3} \right) $ , find the values of $ a $ and $ b $ .

Hint: In this question, we need to determine the values of $ a $ and $ b $ . For this, we will first consider the given that $ \left( {x - 2} \right) $ is a factor of polynomial $ \left( {{x^3} + a{x^2} + bx + 6} \right) $ and use remainder theorem and determine $ p\left( 2 \right) = 0 $ . Here, we will get an equation. Then, we will again consider the given that the polynomial $ \left( {{x^3} + a{x^2} + bx + 6} \right) $ leaves remainder $ 3 $ when divided by $ \left( {x - 3} \right) $ . With this condition we will use remainder theorem and then determine $ p\left( 3 \right) = 3 $ . By which another equation will be formed. By solving both the equations, we will get the required values of $ a $ and $ b $ .

Complete step-by-step answer:
It is given that $ \left( {{x^3} + a{x^2} + bx + 6} \right) $ has $ \left( {x - 2} \right) $ as a factor and leaves remainder $ 3 $ when divided by $ \left( {x - 3} \right) $ .
 We need to find the values of $ a $ and $ b $ .
(i)Let $ p\left( x \right) = \left( {{x^3} + a{x^2} + bx + 6} \right) $
And, $ \left( {x - 2} \right) $ is a factor of polynomial $ \left( {{x^3} + a{x^2} + bx + 6} \right) $ .
As, 2 is the zero of the polynomial $ \left( {{x^3} + a{x^2} + bx + 6} \right) $ , therefore by remainder theorem, we have,
 $ p\left( 2 \right) = 0 $
 $ \left( {{2^3} + a{{.2}^2} + b.2 + 6} \right) = 0 $
 $ 8 + 4a + 2b + 6 = 0 $
 $ 14 + 4a + 2b = 0 $
 $ 4a + 2b = - 14 $
Dividing each term by $ 2 $ , we have,
 $ 2a + b = - 7 $ which is equation 1
(ii) Also, it is given that the polynomial $ \left( {{x^3} + a{x^2} + bx + 6} \right) $ leaves remainder $ 3 $ when divided by $ \left( {x - 3} \right) $ . Therefore, by using remainder theorem,
 $ p\left( 3 \right) = 3 $
 $ {3^3} + a{.3^2} + b.3 + 6 = 3 $
 $ 27 + 9a + 3b + 6 = 3 $
 $ 9a + 3b + 33 = 3 $
 $ 9a + 3b = - 30 $
Dividing each term by 3, we have,
 $ 3a + b = - 10 $ which is equation 2
Now, let us solve equation 1 and 2,
 $ 2a + b = - 7 $
 $ 3a + b = - 10 $
Subtracting equation 2 from 1,
 $ 2a + b - 3a - b = - 7 - \left( { - 10} \right) $
 $ - a = 3 $
Hence, $ a = - 3 $
Substituting the value of $ a $ in equation 1, we have,
 $ 2a + b = - 7 $
 $ \Rightarrow 2\left( { - 3} \right) + b = - 7 $
 $ b = - 7 + 6 $
Hence, $ b = - 1 $
Hence, the values of $ a $ and $ b $ are $ - 3 $ and $ - 1 $ respectively.
So, the correct answer is “the values of $ a $ and $ b $ are $ - 3 $ and $ - 1 $ respectively”.

Note: In this question it is worth noting here that, Remainder theorem is an approach of the Euclidean division of polynomials. It states that the remainder of the division of a polynomial $f\left( x \right)$ by a linear polynomial $x - r$ is equal to $f\left( r \right)$. When we face these kinds of problems we can use the remainder theorem and get the required solution.