Given #{(p(x)=x^4+a x^3+b x^2+c x+1),(q(x)=x^4+c x^3+b x^2+a x + 1):}# find the conditions for #a, b, c, (a ne c)# such that #p(x)# and #q(x)# have two common roots, then solve #p(x)=0# and #q(x) = 0#?

2 Answers
Sep 13, 2016

Answer:

#a+c=0# and #b = -2#

The zeros are #x = +-1#

Explanation:

Given:

#p(x) = x^4+ax^3+bx^2+cx+1#

#q(x) = x^4+cx^3+bx^2+ax+1#

with two common roots and #a != c#

Note that #p(0) = q(0) = 1#, so #x=0# is not a root.

If #x_1# is either of these zeros, then:

#0 = p(x_1) - q(x_1)#

#color(white)(0) = (a-c)x_1^3+(c-a)x_1#

#color(white)(0)= (a-c)x_1(x_1-1)(x_1+1)#

Hence the two roots are #x = +-1#

Then:

#0 = p(1) = a+b+c+2#

#0 = p(-1) = -a+b-c+2#

Adding and subtracting these two equations, we find:

#b = -2#

#a+c = 0#

Sep 13, 2016

Answer:

See bellow.

Explanation:

Given #p(x) = m(x)n_1(x)# and #q(x) = m(x)n_2(x)#
follows that

#p(x)-q(x) = m(x)(n_1(x)-n_2(x))# and

#p(x)-q(x) = (a - c) x (x^2-1)#

so

#n_1(x)-n_2(x)=(a-c)x# because #p(0) = q(0) = 1 ne 0#

and

#m(x) = x^2-1#

now

#p(x) = m(x)(x^2+r_1x-1)# and
#q(x) = m(x)(x^2+r_2x-1)#

Equating for all #x in RR# we obtain the conditions

for #p(x)#
#{(c + r_1=0),(2 + b=0),(a - r_1=0):}#

and for #q(x)#
#{(a + r_2=0),(2 + b=0),(c - r_2=0):}#

So the polynomials are

#p(x) = (x^2-1)(x^2+ax-1)#
#q(x) = (x^2-1)(x^2-ax-1)#