How do you factor #2x^2 -7x -4#?

1 Answer
Oct 31, 2015

Answer:

#2x^2-7x-4 = (2x+1)(x-4)#

Explanation:

Start by hoping to find integer coefficient factors:

#{: ("factors of " x^2 ,color(white)("XX"),"factors of constant",color(white)("XX"),"sum of"), ( "coefficient",color(white)("XX")," coefficient",color(white)("XX"),"cross product"), ((1,2),color(white)("XX"),(-1,4), color(white)("XX"), 1xx4+2xx(-1) = 6), ( ,color(white)("XX"),(1,-4), color(white)("XX"), 1xx(-4)+2xx(-1)=-2), ( ,color(white)("XX"),(-2,2), color(white)("XX"), 1xx2+2xx(-2) = -2), ( ,color(white)("XX"),(2,-2), color(white)("XX"), 1xx(-2)+2xx2=2), ( ,color(white)("XX"),(-4,1), color(white)("XX"), 1xx1+2x(-4)=-7), (,color(white)("XX"),(4,-1),color(white)("XX"), 1xx(-1)+2xx(4)= 7) :}#

The pairs #(1,2) and (-4,1)# give us the middle coefficient that we need.

Therefore the factors are:
#color(white)("XXX")(1x-4)(2x+1)#