# 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?

Sep 13, 2016

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

The zeros are $x = \pm 1$

#### Explanation:

Given:

$p \left(x\right) = {x}^{4} + a {x}^{3} + b {x}^{2} + c x + 1$

$q \left(x\right) = {x}^{4} + c {x}^{3} + b {x}^{2} + a x + 1$

with two common roots and $a \ne c$

Note that $p \left(0\right) = q \left(0\right) = 1$, so $x = 0$ is not a root.

If ${x}_{1}$ is either of these zeros, then:

$0 = p \left({x}_{1}\right) - q \left({x}_{1}\right)$

$\textcolor{w h i t e}{0} = \left(a - c\right) {x}_{1}^{3} + \left(c - a\right) {x}_{1}$

$\textcolor{w h i t e}{0} = \left(a - c\right) {x}_{1} \left({x}_{1} - 1\right) \left({x}_{1} + 1\right)$

Hence the two roots are $x = \pm 1$

Then:

$0 = p \left(1\right) = a + b + c + 2$

$0 = p \left(- 1\right) = - a + b - c + 2$

Adding and subtracting these two equations, we find:

$b = - 2$

$a + c = 0$

Sep 13, 2016

See bellow.

#### Explanation:

Given $p \left(x\right) = m \left(x\right) {n}_{1} \left(x\right)$ and $q \left(x\right) = m \left(x\right) {n}_{2} \left(x\right)$
follows that

$p \left(x\right) - q \left(x\right) = m \left(x\right) \left({n}_{1} \left(x\right) - {n}_{2} \left(x\right)\right)$ and

$p \left(x\right) - q \left(x\right) = \left(a - c\right) x \left({x}^{2} - 1\right)$

so

${n}_{1} \left(x\right) - {n}_{2} \left(x\right) = \left(a - c\right) x$ because $p \left(0\right) = q \left(0\right) = 1 \ne 0$

and

$m \left(x\right) = {x}^{2} - 1$

now

$p \left(x\right) = m \left(x\right) \left({x}^{2} + {r}_{1} x - 1\right)$ and
$q \left(x\right) = m \left(x\right) \left({x}^{2} + {r}_{2} x - 1\right)$

Equating for all $x \in \mathbb{R}$ we obtain the conditions

for $p \left(x\right)$
$\left\{\begin{matrix}c + {r}_{1} = 0 \\ 2 + b = 0 \\ a - {r}_{1} = 0\end{matrix}\right.$

and for $q \left(x\right)$
$\left\{\begin{matrix}a + {r}_{2} = 0 \\ 2 + b = 0 \\ c - {r}_{2} = 0\end{matrix}\right.$

So the polynomials are

$p \left(x\right) = \left({x}^{2} - 1\right) \left({x}^{2} + a x - 1\right)$
$q \left(x\right) = \left({x}^{2} - 1\right) \left({x}^{2} - a x - 1\right)$