# Question #75b0d

May 23, 2017

$y = \left(x - \sqrt{3}\right) \left(x - 4\right) \left({x}^{2} - 4 x + 5\right)$

#### Explanation:

Real roots do not need to be in pairs, even if they are radical expressions. So we already have two parts of our polynomial:

$\left(x - \sqrt{3}\right) \mathmr{and} \left(x - 4\right)$

However, the polynomial as a whole must not contain imaginary numbers, so we must provide a pair for the root $2 - i$ in order to cancel out and make it real.

To do this, all we have to do is make the imaginary part negative (or in this case positive since it is already negative).

Therefore, this polynomial must also have the root $2 + i$.

Finally, multiply the imaginary roots together to make a real polynomial.

$\left(x - \left(2 - i\right)\right) \left(x - \left(2 + i\right)\right)$

$\left(x - 2 + i\right) \left(x - 2 - i\right)$

$\left(\left(x - 2\right) + i\right) \left(\left(x - 2\right) - i\right)$

${\left(x - 2\right)}^{2} - {i}^{2}$

${x}^{2} - 4 x + 4 + 1$

${x}^{2} - 4 x + 5$

So our final polynomial is:

$y = \left(x - \sqrt{3}\right) \left(x - 4\right) \left({x}^{2} - 4 x + 5\right)$