**Example 2:**

Find the roots of the polynomial $x^3 - 2x^2 + 5x + 26$ and hence factor completely.

**Solution:**

The roots of the given polynomial are the solutions of the polynomial equation $x^3 - 2x^2 + 5x + 26$ = $0$

We find there is no scope of grouping the terms to get common factors. Hence let us use the rational root theorem. The possible rational roots are from all possible

combinations of,

$\pm$ $\frac{a\ factor\ of\ 26}{a\ factor\ of\ 1}$

In other words the possible rational roots can be listed as, $\pm 1,\ \pm 2$ and $\pm 13$. Out of these, by trial and error we can see $x$ = $-2$ satisfies the equation $x^3 - 2x^2 + 5x + 26$ = $0$. Hence $-2$ is one of the roots of and hence $(x + 2)$ is a factor of the given polynomial.

The quotient of $\frac{2x^3 - 2x^2 + 5x + 26}{x + 2}$ is $(x^2 -4x + 13)$. Therefore, the remaining roots of the given polynomial are the solutions of the equation $x^2 -4x + 13$ = $0$.

This is a quadratic which cannot be solved by factoring. Hence, let us use the method of completing square.

$x^2 - 4x + 13$ = $0$

$x^2 - 4x$ = $-13$

$x^2 - 4x + 4$ = $-13 + 4$ (Adding $4$ on both sides to complete the square on the left side)

$(x – 2)^2$ = $-9$

So, $x – 2$ = $\pm$ $\sqrt{(-9)}$ = $\pm\ (\sqrt9)(\sqrt{(-1)}$ = $\pm 3i$.

Therefore, $x$ = $2 + 3i$ and $x$ = $2 – 3i$.

Thus, the remaining two roots of the given polynomial are $(2 + 3i)$ and $(2 – 3i)$ and the remaining factors are $(x – 2 – 3i)$ and $(x – 2 +3i)$

So, the given polynomial $x^3 - 2x^2 + 5x + 26$ is completely factored as $(x + 2)(x – 2 – 3i)(x – 2 +3i)$