#(x+a)# is a factor of #f(x)=x^2+px+q# as well as #g(x)=x^2+mx+n#. Prove that #a=(n-q)/(m-p)#?

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Answer 1 · 2017-06-13T05:55:05

Please see below.

We can use here factor theorem, which states that a polynomial #f(x)# has a factor #(x-alpha)#, then #f(alpha)=0#.

Here we are given that #(x+a)# is a factor of #f(x)=x^2+px+q# as well as #g(x)=x^2+mx+n#. Hence we have #f(-a)=0# and #g(-a)=0#, i.e.

#a^2-pa+q=0# .....................(1) and

#a^2-ma+n=0# .....................(2)

Subtracting equation (2) from (1) we get

#-pa+ma+q-n=0#

or #a(m-p)=n-q#

or #a=(n-q)/(m-p)#.