How do you factor the trinomial #4y^4x + 14yx^2 + 10y^2x^2 - 8xy#?

1 Answer
George C.
Dec 12, 2015

#4y^4x+14yx^2+10y^2x^2-8xy=2yx(2y^3+7x+5yx-4)#

with no simpler factorisation.

Explanation:

This is a quadrinomial not a trinomial.

First separate out the common factor #2yx#:

#4y^4x+14yx^2+10y^2x^2-8xy=2yx(2y^3+7x+5yx-4)#

I suspect this is as far as we can factor it, but let's try.

Since the leading term is #2y^3# and all other terms are of lower degree, then if #(2y^3+7x+5yx-4)# splits into two factors then one is linear and the other is quadratic:

#2y^3+7x+5yx-4#

#=(2y+ax+b)(y^2+cyx+dx^2+ey+fx+g)#

#=2y^3+(a+2c)y^2x+adx^3+(b+2e)y^2+(ac+2d)yx^2+(bc+ae+2f)yx+(be+2g)y+(bd+af)x^2+(bf+ag)x+bg#

Hence:

(i) #a+2c=0#

(ii) #ad=0#

(iii) #b+2e=0#

(iv) #ac+2d=0#

(v) #bc+ae+2f=5#

(vi) #be+2g=0#

(vii) #bd+af=0#

(viii) #bf+ag=7#

(ix) #bg=-4#

From (ii) at least one of #a=0# or #d=0#.

If #a=0# then from (iv) we find #d=0#.

So we at least know #d=0#

Since #d=0#, then from (iv) we find #ac=0#, so at least one of #a=0# or #c=0#. Then from (i) we find that the other is zero too.

So given that #a = c = d = 0#, the remaining equations become:

(iii) #b+2e=0#

(v) #2f=5#

(vi) #be+2g=0#

(viii) #bf=7#

(ix) #bg=-4#

From (v) #f = 5/2#

From (viii) #b = 7/f = 14/5#

From (ix) #g = -4/b = -20/14 = -10/7#

From (iii) #e = -b/2 = -7/5#

So:

#be+2g = 14/5*(-7/5)+2(-10/7)#

#= -98/25-20/7 != 0# violating (vi)

So there is no simpler factorisation.

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