# What are the possible solutions of ax^4+bx^3+cx^2+dx+e=0 where a!=0, a+c+e=0 and b+d=0?

Mar 22, 2018

This quartic has roots $x = \pm 1$ and:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a \left(a + c\right)}}{2 a}$

#### Explanation:

Let:

$f \left(x\right) = a {x}^{4} + b {x}^{3} + c {x}^{2} + \mathrm{dx} + e$

with $a \ne 0$, $a + c + e = 0$ and $b + d = 0$

Then:

$f \left(1\right) = a + b + c + d + e = \left(a + c + e\right) + \left(b + d\right) = 0$

$f \left(- 1\right) = a - b + c - d + e = \left(a + c + e\right) - \left(b + d\right) = 0$

So $x = 1$ and $x = - 1$ are zeros with corresponding factors $\left(x - 1\right)$ and $\left(x + 1\right)$.

Hence $f \left(x\right)$ is divisible by:

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

We have:

$a {x}^{4} + b {x}^{3} + c {x}^{2} + \mathrm{dx} + e = \left({x}^{2} - 1\right) \left(a {x}^{2} + b x + \left(a + c\right)\right)$

Note that from the given conditions on $a , b , c , d , e$ there are no restrictions on $a , b , c$ apart from $a \ne 0$.

So the remaining quadratic can have any coefficients.

We can express the solutions using the quadratic formula, substituting $\left(a + c\right)$ for "c" to find zeros:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a \left(a + c\right)}}{2 a}$