Question

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`?

Answer 1

`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`

Answer 2

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)`