# Question #27939

Aug 2, 2015

As Sudip Sinha has pointed out $- 1 + \sqrt{3} i$ is NOT a zero. (I neglected to check that.) The other zeros are $1 - \sqrt{3} i$ and $1$.

#### Explanation:

Because all of the coefficients are real numbers, any imaginary zeros must occur in conjugate pairs.
Therefore, $1 - \sqrt{3} i$ is a zero.

If $c$ is a zero then $z - c$ is a factor, so we could multiply

$\left(z - \left(1 + \sqrt{3} i\right)\right) \left(z - \left(1 - \sqrt{3} i\right)\right)$ to get ${z}^{2} - 2 z + 4$
and then divide $P \left(z\right)$ by that quadratic.

But it's quicker to consider the possible rational zero for $P$ first. Or add the coefficients to see that $1$ is also a zero.

Aug 2, 2015

$1$ and $1 - \sqrt{3} i$

#### Explanation:

There is an error in your question. The root should be $1 + \sqrt{3} i$. You can verify this by putting the value in the expression. If it is a root the expression should evaluate to zero.

The expression has all real coefficients, so by the Complex Conjugate Roots Theorem (https://en.wikipedia.org/wiki/Complex_conjugate_root_theorem), we have that the other complex root is $1 - \sqrt{3} i$,

Clearly, the third root (say $a$) has to be real, since it cannot have a complex conjugate; otherwise there will be 4 roots, which is not possible for a 3rd degree equation.

Note
$\left(z - \left(1 - \sqrt{3} i\right)\right) \left(z - \left(1 + \sqrt{3} i\right)\right)$
$= \left(\left(z - 1\right) + \sqrt{3} i\right) \left(\left(z - 1\right) - \sqrt{3} i\right)$
$= \left({\left(z - 1\right)}^{2} - {\left(\sqrt{3} i\right)}^{2}\right)$ ( Since $\left(z + a\right) \left(z - a\right) = {z}^{2} - {a}^{2}$.)
$= {z}^{2} - 2 z + 1 - 3 \left(- 1\right)$
$= {z}^{2} - 2 z + 4$

We will try to get this factor in the expression.
We may write:
$P \left(z\right) = {z}^{3} - 3 {z}^{2} + 6 z - 4$
$= z \left({z}^{2} - 2 z + 4\right) - 1 \left({z}^{2} - 2 z + 4\right)$
$= \left(z - 1\right) \left({z}^{2} - 2 z + 4\right)$
$= \left(z - 1\right) \left(z - \left(1 - \sqrt{3} i\right)\right) \left(z - \left(1 + \sqrt{3} i\right)\right)$

Aug 2, 2015

As an intro, I think that the root should be $\textcolor{b l u e}{1 + \sqrt{3}}$ and not $\textcolor{red}{- 1 + \sqrt{3}}$

On that basis my answer is :

$z \in \left\{1 , \text{ "1+sqrt3," } 1 - \sqrt{3}\right\}$

#### Explanation:

By using the idea of complex conjugates and some other cool tricks.

$P \left(z\right)$ is a polynomial of degree $3$. This implies that it should only have $3$ roots.

One interesting fact about complex roots is that they never occur alone.They always occur in conjugate pairs.

So if $1 + i \sqrt{3}$ is one root, then its conjugate : $1 - i \sqrt{3}$ most certainly is a root too!

And since there is just one more root left, we can call that root $z = a$.
It is not a complex number because, complex roots always occur in pairs.
And since this is the last of the $3$ roots, there cannot be any other pair after the first one!

In the end the factors of $P \left(z\right)$ were easily found to be $\left[z - \left(1 + i \sqrt{3}\right)\right] \text{ , " [z-(1-isqrt3)] " and } \left(z - a\right)$

NB : Note that the difference between a root and a factor is that :
- A root could be $z = 1 + i$
But the corresponding factor would be $z - \left(1 + i\right)$

The second trick is that, by factoring $P \left(z\right)$ we should get something like this :

$P \left(z\right) = \left[z - \left(1 + i \sqrt{3}\right)\right] \left[z - \left(1 - i \sqrt{3}\right)\right] \left(z - a\right)$

Next, expand the braces,

$P \left(z\right) = \left[{z}^{2} - z \left(1 + i \sqrt{3} + 1 - i \sqrt{3}\right) + \left(1 + i \sqrt{3}\right) \left(1 - i \sqrt{3}\right)\right] \left(z - a\right)$

$= \left[{z}^{2} - z \left(2\right) + \left(1 + 3\right)\right] \left(z - a\right)$

$= \left[{z}^{2} - 2 z + 4\right] \left(z - a\right)$

$= {z}^{3} + {z}^{2} \left(- a - 2\right) + z \left(2 a + 4\right) - 4 a$

Next, we equate this to the original polynomial $P \left(z\right) = {z}^{3} - 3 {z}^{2} + 6 z - 4$

$\implies {z}^{3} + {z}^{2} \left(- a + 2\right) + z \left(- 2 a + 4\right) - 4 a = {z}^{3} - 3 {z}^{2} + 6 z - 4$

Since the two polynomials are identical, we equate the coefficients of ${z}^{3}$, ${z}^{2}$, ${z}^{1}$and ${z}^{0}$(the constant term) on either side,

Actually,we just need to pick one equation and to solve it for $a$

Equating the constant terms,

$\implies - 4 a = - 4$

$\implies a = 1$

Hence the last root is $\textcolor{b l u e}{z = 1}$