Question 101829: For what integral values of k can 4x^2+kx+3 be factored over the integers?
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website! The possible factors of this quadratic are as follows:
(2x+1)(2x+3)
(4x+1)(x+3)
(4x+3)(x+1)
We will use the FOIL crutch (First, Inner, Outer, Last) to expand the above:
(2x+1)(2x+3)----4x^2+2x+6x+3=4x^2+8x+3;k=8
(4x+1)(x+3)-----4x^2+x+12x+3=4x^2+13x+3; k=13
(4x+3)(x+1)-----4x^2+3x+4x+3=4x^2+7x+3; k=7
So the answer is:
k=7, 8 or 13
There may be a more scientific approach to solving this problem, but if there is I'm not aware of it. Generally, when the A coefficient of a quadratic is >1, some level of trial and error is needed to get at the factors. There are some tricks, however, that simplifies the trial and error process. In this quadratic, for example, since all signs are positive then we know there cannot be any negative signs in the factors. Also, if the C term is positive and the B term is negative,then we know, right off, that both signs in each of the two factors must be negative. If both the B term and the C term are negative, then we know that one factor is positive and one is negative. Information such as this can frequently help to reduce the number of possibilities.
Hope this helps----ptaylor
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