# How do you factor x^3 + 7x^2 + 9x + 3?

Jul 26, 2015

Use shortcut concerning coefficients, then quadratic formula to find:

${x}^{3} + 7 {x}^{2} + 9 x + 3 = \left(x + 1\right) \left(x + 3 + \sqrt{6}\right) \left(x + 3 - \sqrt{6}\right)$

#### Explanation:

Let $f \left(x\right) = {x}^{3} + 7 {x}^{2} + 9 x + 3$

First notice that the sum of coefficients of the terms of odd powers of $x$ is equal to the sum of the coefficients of the terms of even powers of $x$. That is $1 + 9 = 10 = 7 + 3$.

As a result, $x = - 1$ is a root of $f \left(x\right) = 0$

$f \left(- 1\right) = - 1 + 7 - 9 + 3 = 0$

So $\left(x + 1\right)$ is a factor of $f \left(x\right)$

${x}^{3} + 7 {x}^{2} + 9 x + 3 = \left(x + 1\right) \left({x}^{2} + 6 x + 3\right)$

Next ${x}^{2} + 6 x + 3$ is of the form $a {x}^{2} + b x + c$, with $a = 1$, $b = 6$ and $c = 3$. This has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {6}^{2} - \left(4 \times 1 \times 3\right) = 36 - 12 = 24$

Since this is positive, but not a perfect square ${x}^{2} + 6 x + 3 = 0$ has irrational roots given by the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{- b \pm \sqrt{\Delta}}{2 a}$

$= \frac{- 6 \pm \sqrt{24}}{2} = - 3 \pm \sqrt{6}$

Hence ${x}^{2} + 6 x + 3 = \left(x + 3 + \sqrt{6}\right) \left(x + 3 - \sqrt{6}\right)$

Putting it all together:

${x}^{3} + 7 {x}^{2} + 9 x + 3 = \left(x + 1\right) \left(x + 3 + \sqrt{6}\right) \left(x + 3 - \sqrt{6}\right)$