How do you factor #1 - 4b² + a² - 2ab#?

Question

1443 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-27T20:32:09

This expression does not factorise.

Given: #1-4b^2+a^2-2ab#

Notice that the expression is a mixture of terms of degree #2# and degree #0#.

If it did factor then the factors would be a mixture of terms of degree #1# and #0#.

The factors can be scaled so that the coefficient of #a# is #1# in both factors, resulting in a factorisation of the form:

#1-4b^2+a^2-2ab#

#=(a+pb+r)(a+qb+s)#

#=a^2+(p+q)ab+pqb^2+(r+s)a+(ps+qr)b+rs#

Equating coefficients, we find:

#{(p+q=-2), (pq=-4), (r+s=0), (ps+qr=0), (rs=1) :}#

From the first two equations, we find that #p# and #q# are (in some order):

#{ (-1+sqrt(5)), (-1-sqrt(5)) :}#

From the third and fifth equation, we find that #r# and #s# are (in some order):

#{ (i), (-i) :}#

Then we find that:

#ps+qr = +-2sqrt(5)i#

contradicting the fourth equation.

So there is no such factorisation.