# How do you construct polynomial equations with the given roots?

## 1. $2$, $4$ and $- 7$. 2. $5$ and $\sqrt{3}$.

Aug 11, 2017

1. ${x}^{3} + {x}^{2} - 34 x + 56 = 0$

2. ${x}^{3} - 5 {x}^{2} - 3 x + 15 = 0$

#### Explanation:

Note that if a polynomial in $x$ has a zero $a$ then it has a factor $\left(x - a\right)$ and vice versa.

$\textcolor{w h i t e}{}$
For question 1 we can construct a polynomial:

$f \left(x\right) = \left(x - 2\right) \left(x - 4\right) \left(x + 7\right) = {x}^{3} + {x}^{2} - 34 x + 56$

Any polynomial with these zeros will be a multiple (scalar or polynomial) of this $f \left(x\right)$.

So the polynomial equation:

${x}^{3} + {x}^{2} - 34 x + 56 = 0$

has roots $2$, $4$ and $- 7$.

$\textcolor{w h i t e}{}$
For question 2 we can multiply out $\left(x - 5\right) \left(x - \sqrt{3}\right)$ but this will result in a polynomial with irrational coefficients:

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

If - as is probably the case - we want a polynomial with integer coefficients, then we also need the rational conjugate $- \sqrt{3}$ to be a zero and $\left(x + \sqrt{3}\right)$ a factor.

Then we can define:

$g \left(x\right) = \left(x - 5\right) \left(x - \sqrt{3}\right) \left(x + \sqrt{3}\right) = \left(x - 5\right) \left({x}^{2} - 3\right) = {x}^{3} - 5 {x}^{2} - 3 x + 15$

Any polynomial with these zeros will be a multiple (scalar or polynomial) of this $g \left(x\right)$.

So the polynomial equation:

${x}^{3} - 5 {x}^{2} - 3 x + 15 = 0$

has roots $5$, $\sqrt{3}$ and $- \sqrt{3}$.