# How do you write the polynomial function with the least degree and zeroes 1, 4, 1 + √2?

Mar 5, 2016

#### Answer:

${x}^{3} - \left(6 + \sqrt{2}\right) {x}^{2} + \left(9 + 5 \sqrt{2}\right) x - 4 \left(1 + \sqrt{2}\right)$

#### Explanation:

A polynomial function with the least degree and zeroes $\left\{1 , 4 , \left(1 + \sqrt{2}\right)\right\}$ will be given by

$\left(x - 1\right) \left(x - 4\right) \left(x - \left(1 + \sqrt{2}\right)\right)$ or

$\left(x - 1 - \sqrt{2}\right) \left[x \left(x - 4\right) - 1 \left(x - 4\right)\right]$ or

$\left(x - 1 - \sqrt{2}\right) \left({x}^{2} - 4 x - x + 4\right)$ or $\left(x - 1 - \sqrt{2}\right) \left({x}^{2} - 5 x + 4\right)$ or

$x \left({x}^{2} - 5 x + 4\right) - 1 \left({x}^{2} - 5 x + 4\right) - \sqrt{2} \left({x}^{2} - 5 x + 4\right)$ or

${x}^{3} - 5 {x}^{2} + 4 x - {x}^{2} + 5 x - 4 - \sqrt{2} {x}^{2} + 5 \sqrt{2} x - 4 \sqrt{2}$ or

${x}^{3} - \left(6 + \sqrt{2}\right) {x}^{2} + \left(9 + 5 \sqrt{2}\right) x - 4 \left(1 + \sqrt{2}\right)$