Question 1174668
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            According to the context, the coefficients of linear binomials are (should be) integer numbers.


            Otherwise,  the problem has  INFINITELY  MANY  solutions.



<pre>
Consider 

    10x^2 + tx + 8 = (ax+b)*(cx+d).


Then we want to have  ac = 10,  bd = 8.


We have these cases


    a= 10, c= 1,  b= 8, d= 1;   (10x+8)*(1x+1) = 10x^2 + 18x + 8;   t = 18.

    a=  5, c= 2,  b= 8, d= 1;   ( 5x+8)*(2x+1) = 10x^2 + 21x + 8;   t = 21.



    a= -10, c= -1,  b= 8, d= 1;   (-10x+8)*(-1x+1) = 10x^2 - 18x + 8;  t = -18.

    a=  -5, c= -2,  b= 8, d= 1;   ( -5x+8)*(-2x+1) = 10x^2 + 21x + 8;  t = -21.
</pre>


It is only the beginning of an analysis.


The full number of cases is &nbsp;MUCH &nbsp;MORE.


I showed the idea to you.


Having a patience, &nbsp;you may continue and complete it &nbsp;ON &nbsp;YOUR &nbsp;OWN.



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;REMEMBER:  &nbsp;you must satisfy two equalities  &nbsp;&nbsp;ac= 10,  &nbsp;&nbsp;bd= 8  &nbsp;in integer  numbers by trial and error; 


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;try all appropriate positive and negative integer pairs.  &nbsp;Then for &nbsp;"t" &nbsp;you have this values  &nbsp;&nbsp;t = ad + bc.



I wish you pleasant pastime &nbsp;(!)



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Ignore the post by &nbsp;@josgarithmetic, &nbsp;since it is &nbsp;IRRELEVANT &nbsp;to the posed problem.



Do not distract yourself reading the post from  @MathLover1.



Learn from good sources (!) (!)