How do you factor x^3-9x^2+25x-21, given that x=3 is a zero ?

Nov 2, 2016

${x}^{3} - 9 {x}^{2} + 25 x - 21 = \left(x - 3\right) \left(x - 3 - \sqrt{2}\right) \left(x - 3 + \sqrt{2}\right)$

Explanation:

$h \left(x\right) = {x}^{3} - 9 {x}^{2} + 25 x - 21$

The difference of squares identity can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

We use this later with $a = \left(x - 3\right)$ and $b = \sqrt{2}$

We are told that $3$ is a zero, so $\left(x - 3\right)$ must be a factor:

${x}^{3} - 9 {x}^{2} + 25 x - 21 = \left(x - 3\right) \left({x}^{2} - 6 x + 7\right)$

Then, completing the square we find:

${x}^{2} - 6 x + 7 = {x}^{2} - 6 x + 9 - 2$

$\textcolor{w h i t e}{{x}^{2} - 6 x + 7} = {\left(x - 3\right)}^{2} - {\left(\sqrt{2}\right)}^{2}$

$\textcolor{w h i t e}{{x}^{2} - 6 x + 7} = \left(\left(x - 3\right) - \sqrt{2}\right) \left(\left(x - 3\right) + \sqrt{2}\right)$

$\textcolor{w h i t e}{{x}^{2} - 6 x + 7} = \left(x - 3 - \sqrt{2}\right) \left(x - 3 + \sqrt{2}\right)$

Putting it all together:

${x}^{3} - 9 {x}^{2} + 25 x - 21 = \left(x - 3\right) \left(x - 3 - \sqrt{2}\right) \left(x - 3 + \sqrt{2}\right)$