Chapter 8.6, Problem 48HP

To determine

To find: All the values of k so that the polynomial can be factored using integers.

Answer to Problem 48HP

  $2,3$

Explanation of Solution

Given:

  $2x^2-5x+k,k>0$

Calculation:

The given polynomial is $2x^2-5x+k,k>0$

Comparing this with $a{x}^2+bx+c$ ,

  $a=2,b=-5,c=k$

The value of $ac$ is $2k$

From ac method the middle term b should be equal to the sum/difference of factors pairs of ac .

So, $-5$ can be written as

  $\begin{array}{l}-5=-1-4\\ -5=-2-3\end{array}$

The value of ac should be the product of these numbers. So, the values of ac are

  $\begin{array}{l}\left(-1\right)\left(-4\right)=4\\ \left(-2\right)\left(-3\right)=6\end{array}$

These values should be equal to $2k$

Hence,

  $\begin{array}{l}2k=4\Rightarrow k=2\\ and\\ 2k=6\Rightarrow k=3\end{array}$

Hence, the values of k are $2,3$