Finding solution of an algebraic equation by Horner's synthetic division method using a complex root

Question

If $(2+3i)$ is a root of $x^4-x^3+3x^2+31x+26=0$, then it is easy to find the roots of the given equation using usual long division method. My question is, 'will it be possible to use Horner's synthetic division method instead of usual long division method to reduce the given equation (i.e., divide the given polynomial by $(x-(2+3i))$ and $(x-(2-3i))$) successively'?

Answer 1

Another way is to factor the polynomial, i.e., $$ x^4-x^3+3x^2+31x+26=(x^2 - 4x + 13)(x + 2)(x + 1). $$ We see that $-2$ and $-1$ are roots by the rational root test, and then get a quadratic equation using division by $x+1$ and $x+2$. The quadratic equation $x^2-4x+13$ has the complex roots $2\pm 3i$.

Answer 2

By the Ferrari's method $$x^4-x^3+3x^2+31x+26=\left(x^2-\frac{1}{2}x+\frac{15}{2}\right)^2-\left(\frac{7}{2}x-\frac{11}{2}\right)^2=$$ $$=(x^2-4x+13)(x^2+3x+2).$$

Answer 3

Alt. hint: $\;2-3i\,$ must be a root as well, then let the other two roots be $\,a,b\,$. By Vieta's relations:

$$\require{cancel} (2+\cancel{3i})+(2-\cancel{3i}) + a + b = 1 \quad\iff\quad a+b = 1 - 4 = -3 \\ (2+3i) \cdot (2-3i) \cdot a \cdot b = 26 \quad\iff\quad ab = 26 / 13 = 2 $$

It follows that $\,a,b\,$ are the roots of the quadratic $\,x^2+3x+2 = (x+1)(x+2)\,$.