# How do you write a polynomial with zeros: 2, 4 + sqrt5, 4-sqrt5?

Feb 9, 2016

${x}^{4} - 14 {x}^{3} + 67 {x}^{2} - 130 x + 88$

#### Explanation:

Note that: if $a$ is a zero of a polynomial, then $\left(x - a\right)$ is a root of the polynomial. Applying this to the zeros at hand, we see that that polynomial is equal to

$\left(x - 2\right) \left(x - 4\right) \left(x - \left(4 + \sqrt{5}\right)\right) \left(x - \left(4 - \sqrt{5}\right)\right)$

We can simplify the first two through distributing. We will get to the second two, which is the more challenging part of this problem, but first distribute the minus sign.

$= \left({x}^{2} - 6 x + 8\right) \left(x - 4 - \sqrt{5}\right) \left(x - 4 + \sqrt{5}\right)$

$= \left({x}^{2} - 6 x + 8\right) \left(\left(x - 4\right) - \sqrt{5}\right) \left(\left(x - 4\right) + \sqrt{5}\right)$

Notice that the last two terms are in the form $\left(a - b\right) \left(a + b\right)$, which equals ${a}^{2} - {b}^{2}$. Here, $a = x - 4$ and $b = \sqrt{5}$. This gives us

$= \left({x}^{2} - 6 x + 8\right) \left({\left(x - 4\right)}^{2} - 5\right)$

$= \left({x}^{2} - 6 x + 8\right) \left({x}^{2} - 8 x + 16 - 5\right)$

$= \left({x}^{2} - 6 x + 8\right) \left({x}^{2} - 8 x + 11\right)$

Now, we can distribute these terms. There will be a lot, so hang on tight.

$= {\overbrace{{x}^{4} - 6 {x}^{3} + 8 {x}^{2}}}^{\left({x}^{2} - 6 x + 8\right) \cdot {x}^{2}} + {\overbrace{- 8 {x}^{3} + 48 {x}^{2} - 64 x}}^{\left({x}^{2} - 6 x + 8\right) \cdot - 8 x} + {\overbrace{11 {x}^{2} - 66 x + 88}}^{\left({x}^{2} - 6 x + 8\right) \cdot 11}$

Combining all the like terms ${x}^{3}$ with ${x}^{3}$, and so on, our result is

$= {x}^{4} - 14 {x}^{3} + 67 {x}^{2} - 130 x + 88$

Note that this polynomial can be multiplied by any constant, and the roots will be the same. However, this is the simplest form of the polynomial.