# How do you write a polynomial in standard form given the zeros +1, -1 , 3 double root?

May 9, 2016

The biquadratic equation with roots $1. - 1 , 3 \mathmr{and} 3$ is
${x}^{4} - 6 {x}^{3} + 8 {x}^{2} + 6 x - 9 = 0$

#### Explanation:

The biquadratic having factors $\left(x - 1\right) , \left(x + 1\right) \left(x + 3\right) \left(x + 3\right)$ is

$f \left(x\right) = \left({x}^{2} - 1\right) \left({x}^{2} - 6 x + 9\right) = {x}^{4} - 6 {x}^{3} + 8 {x}^{2} + 6 x - 9$

So, the equation with roots 1, -1 anmd 3 double root is

$f \left(x\right) = \left({x}^{2} - 1\right) \left({x}^{2} - 6 x + 9\right) = {x}^{4} - 6 {x}^{3} + 8 {x}^{2} + 6 x - 9$