Answer:
[tex]4x^2+xy-18y^2=(4x+9y)\,(x-2y)[/tex]
Step-by-step explanation:
Let's examine the following general product of two binomials with variables x and y in different terms:
[tex](ax+by)\,(cx+dy)= ac\,x^2+adxy+bcxy+bdy^2[/tex]
so we want the following to happen:
[tex]a\,c = 4\\ad+bc=1\\bd--18[/tex]
Notice as well that [tex]ad+bc =1[/tex] means that those two products must differ in just one unit so, one of them has to be negative, or three of them negative. Given that the product [tex]bd=-18[/tex], then we can consider the case in which one of this (b or d) is the negative factor. So let's then assume that [tex]a\,\,and \,\,c[/tex] are positive.
We can then try combinations for [tex]a\,\,and \,\,c[/tex] such as:
[tex]a = 4;\,\,c=1\\a=2;\,\,c=2\\a=1;\,\,c=4[/tex]
Just by selecting the first one [tex](a=4;\,\,c=1)[/tex]
we get that [tex]4d_b=1\\b=-4d-1[/tex]
and since
[tex]bd=-18\\(-4d-1)\,d=-18\\-4d^2-d=-18\\4d^2+d-18=0[/tex]
This quadratic equation give as one of its solutions the integer: d = -2, and consequently,
[tex]d=-18/(-2)\\d=9[/tex]
Now we have a good combination of parameters to render the factoring form of the original trinomial:
[tex]a=4; \,\,b=9;\,\,c=1;\,\,d=-2[/tex]
which makes our factorization:
[tex](4x+9y)\,(x-2y)=4x^2+xy-18y^2[/tex]
