Question

Show work and Factor 4x^2+xy-18y^2

Answer 1

Answer:

$4x^2+xy-18y^2=(4x+9y)\,(x-2y)$

Step-by-step explanation:

Let's examine the following general product of two binomials with variables x and y in different terms:

$(ax+by)\,(cx+dy)= ac\,x^2+adxy+bcxy+bdy^2$

so we want the following to happen:

$a\,c = 4\\ad+bc=1\\bd--18$

Notice as well that $ad+bc =1$ 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 $bd=-18$, then we can consider the case in which one of this  (b or d) is the negative factor. So let's then assume that $a\,\,and \,\,c$ are positive.

We can then try combinations for  $a\,\,and \,\,c$  such as:

$a = 4;\,\,c=1\\a=2;\,\,c=2\\a=1;\,\,c=4$

Just by selecting the first one $(a=4;\,\,c=1)$

we get that $4d_b=1\\b=-4d-1$

and since

$bd=-18\\(-4d-1)\,d=-18\\-4d^2-d=-18\\4d^2+d-18=0$

This quadratic equation give as one of its solutions the integer: d = -2, and consequently,

$d=-18/(-2)\\d=9$

Now we have a good combination of parameters to render the factoring form of the original trinomial:

$a=4; \,\,b=9;\,\,c=1;\,\,d=-2$

which makes our factorization:

$(4x+9y)\,(x-2y)=4x^2+xy-18y^2$