what are the factors of x^2+4xy-x+3y^2+y-2

Question

factoring by regrouping

Answer 1

x²+4xy-x+3y²+y-2=(x+3y-2)(x+y+1)

METHOD

Assume we can write the factors as:

(x+ay+b)(x+cy+d)=x²+cxy+dx+axy+acy²+ady+bx+bcy+bd, where a, b, c, d are constants. We have to find these constants.

Equate coefficients:

constant: bd=-2, d=-2/b;

x: b+d=-1, d=-(b+1); (4)

y: ad+bc=1; d=(1-bc)/a (3)

xy: a+c=4, c=4-a; (1)

y²: ac=3, c=3/a; (2)

Therefore from (1) and (2), c=4-a=3/a, so 4a-a²=3, a²-4a+3=0=(a-1)(a-3), so a=1 or 3⇒c=3 or 1 respectively.

Also, from (3) and (4), d=-b-1=1-3b⇒b=1, d=-2 when a=1 and c=3; or d=-b-1=(1-b)/3, -3b-3=1-b⇒b=-2, d=1 when a=3 and c=1.

Therefore we have (a,b,c,d)=(1,1,3,-2) or (3,-2,1,1). But these yield the same factors x+y+1 and x+3y-2, hence:

x²+4xy-x+3y²+y-2=(x+3y-2)(x+y+1).