Answer

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**Hint**: In the first part factorize the given quadratic polynomial to write it as product of two first degree polynomials and in the second part factorize the polynomial and check whether for what value of $ k $ the equation can be factorized.

**:**

__Complete step-by-step answer__A.The polynomial is given as $ {x^2} - 3x + 2 $ .

As the given polynomial $ {x^2} - 3x + 2 $ is a quadratic polynomial. So, factorize the polynomial by breaking the $ - 3x $ term in two parts such that their coefficients multiplication is equal to $ 2 $ and addition is equal to $ - 3 $ .

So, factorize the given polynomial as shown below:

$

\Rightarrow {x^2} - 3x + 2 = {x^2} - x - 2x + 2 \\

= x\left( {x - 1} \right) - 2\left( {x - 1} \right) \\

= \left( {x - 2} \right)\left( {x - 1} \right) \;

$

So, the polynomial $ {x^2} - 3x + 2 $ can be written as $ \left( {x - 2} \right)\left( {x - 1} \right) $ in the form of product of two first degree polynomials.

**So, the correct answer is “ $ \left( {x - 2} \right)\left( {x - 1} \right) $ ”.**

B.Now in the second part factorize the given polynomial:

$

\Rightarrow {x^2} - 3x + k = {x^2} - 2x - x + k \\

= x\left( {x - 2} \right) - 1\left( {x - k} \right) \;

$

So, this polynomial can only be further written as product of two first degree polynomials if $ x - 2 = x - k $

$\Rightarrow k = 2 $

So, the maximum value of $ k $ for which the polynomial $ {x^2} - 3x + k $ can be written as a product of two first degree polynomials is equal to $ 2 $ .

**So, the correct answer is “k=2”.**

**Note**: In the first part the polynomial is given and need to write it as a product of two first degree polynomials but factorizing it. In the next part a polynomial contains a value as a constant which can only be found with the given condition that the polynomial can be written as a product of two first degree polynomials.

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