Question

Algebra 2

Please see the image below for my question!

Answer 1

Answer:

$x=\pm i,x=-13$

Step-by-step explanation:

Since $x=-1$ is a solution of $x^4+14x^3+14x^2+14x+13$, by the factor theorem, $(x+1)$ is a factor.

We perform the long division as shown in the attachment to obtain:

$x^4+14x^3+14x^2+14x+13=(x+1)(x^3+13x+x+13)$

We now factor the cubic polynomial by grouping.

$x^4+14x^3+14x^2+14x+13=(x+1)(x^2(x+13)+1(x+13))$

$x^4+14x^3+14x^2+14x+13=(x+1)(x^2+1)(x+13)$

Hence the other roots are:

$x=\pm i,x=-13$