Question 1174668
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{{{10x^2+tx+8 = (ax+b)(cx+d)}}}<br>
Since the leading coefficient and constant term are both positive, the coefficients of a and c can be either both positive or both negative; likewise for b and d.<br>
But allowing both pairs to be either positive or negative doesn't provide any different linear terms in the product.  For example...<br>
{{{(2x+3)(3x-1) = 6x^2+7x-3}}} and {{{(-2x-3)(-3x+1) = 6x^2+7x-3}}}<br>
So, for determining the number of possible values of the linear coefficient t, we can assume coefficients a and c are positive.  Then the possible factorizations with integer coefficients have...
(1) (a,c) =(10,1) or (5,2) or (2,5) or (1,10); and
(2) (b,d) = (8,1) or (4,2) or (2,4) or (1,8) OR (-8,-1) or (-4,-2) or (-2,-4) or (-1,-8)<br>
Every combination of choices from (1) and (2) will produce a different value for t.<br>
If the question asked only for the NUMBER of different possible values for t, we would know the answer at this point: 4 choices for the pair of leading coefficients and 8 choices for the pair of constants, so 4*8=32 different values of t.<br>
However, the question asks for a list of all the possible values of t; so there is still some work for you to do to finish the problem.<br>
Note that in making the complete list of the possible values of t, you can use only the positive pairs of values for the constants; whatever (positive) value of t you get, the corresponding negative value of t can be obtained using the corresponding pair of negative constants.<br>
So determine the t values using the positive choices for constants b and d to get 16 of the 32 possible values for t.  Then the remaining 16 possible values of t are the corresponding negative values.<br>