# Factor x^4 - 2x^3 + 4x^2 - 6x + 3 into linear and quadratic terms knowing that -isqrt(3) is a zero?

## I think synthetic division might help. And also this is the real question, ignore the other question that looks very similar. Thanks.

Dec 9, 2017

$\left(x + i \sqrt{3}\right) \left(x - i \sqrt{3}\right) {\left(x - 1\right)}^{2}$

#### Explanation:

If $x = - i \sqrt{3}$ is zero of this polynomial, $x = i \sqrt{3}$ also zero of it. Hence $\left({x}^{2} + 3\right)$ is multiplier of it.

${x}^{4} - 2 {x}^{3} + 4 {x}^{2} - 6 x + 3$

=${x}^{4} + 3 {x}^{2} - 2 {x}^{3} - 6 x + {x}^{2} + 3$

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

=$\left({x}^{2} + 3\right) \cdot \left({x}^{2} - 2 x + 1\right)$

=$\left(x + i \sqrt{3}\right) \left(x - i \sqrt{3}\right) {\left(x - 1\right)}^{2}$