Chapter 8.6, Problem 46HP

To determine

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

Answer to Problem 46HP

  $-25,-14,-11,-10,10,11,14,25$

Explanation of Solution

Given:

  $2x^2+kx+12$

Calculation:

The given polynomial is $2x^2+kx+12$

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

  $a=2,b=k,c=12$

The value of $ac$ is $24$

Now, the factor pairs of 24 are

  $\begin{array}{l}1,24\\ 2,12\\ 3,8,\\ 6,4\end{array}$

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

So, the sum/difference of above factor pairs which is equal to ac are

  $\begin{array}{l}-1-24=-25\\ 1+24=25\\ -2-12=-14\\ 2+12=14\\ -3-8=-11\\ 3+8=11\\ -6-4=-10\\ 6+4=10\end{array}$

Hence, the values of k are $-25,-14,-11,-10,10,11,14,25$