Factorise #3x^2 + xy - 2y^2# ?

2 Answers
Jim G.
Feb 5, 2018

#(3x-2y)(x+y)#

Explanation:

#"using the a-c method"#

#"the factors of - 6 which sum to + 1 are + 3 and - 2"#

#"splitting the middle term "#

#3x^2+3xy-2xy-2y^2larrcolor(blue)"factorise in groups"#

#=color(red)(3x)(x+y)color(red)(-2y)(x+y)#

#"take out the common factor of "(x+y)#

#=(x+y)(color(red)(3x-2y))#

#rArr3x^2+xy-2y^2=(x+y)(3x-2y)#

Douglas K.
Feb 5, 2018

#3x^2 + xy - 2y^2 = (3x-2y)(x+y)#

Explanation:

If #3x^2 + xy - 2y^2# factors, we know that this is the starting form:

#(3x + ay)(x +by)#

Because this makes the product of the first terms become #3x^2#

Multiplying the outside and inside terms we obtain the equation:

#3bxy+axy = xy#

#3b +a = 1" [1]"#

Multiply the last terms:

#aby^2 = -2y^2#

#ab = -2#

#a = -2/b

Substitute into equation [1]:

#3b -2/b = 1#

#3b^2 -2 = b#

#3b^2 - b -2 = 0#

There are two roots:

#b = 1# or #b=-2/3#

The latter does not make sense.

Returning to the equation #ab = -2

#a(1) = -2#

#a = -2#

The factors are:

#(3x-2y)(x+y)#

Related questions
Impact of this question
4444 views around the world
You can reuse this answer
Creative Commons License