# How do you factor x^3+6x^2+11x+6?

Aug 2, 2017

${x}^{3} + 6 {x}^{2} + 11 x + 6 = \textcolor{red}{\left(x + 1\right) \left(x + 2\right) \left(x + 3\right)}$

#### Explanation:

There are several ways to approach this.
One of the most reliable is to hope that the expression has rational roots and apply the Rational Root Theorem.

In this case, the Rational Root Theorem tells us that (if the expression has rational roots) those roots are integer factors of $6$ (the constant term of the expression).

We can build a table/spreadsheet to evaluate the expression for these possible factors: Note that when $x = - 1 \mathmr{and} x = - 2 \mathmr{and} x = - 3$ then the expression has a value of $0$.

Remember that if an expression has a value of $0$ when $x = a$
then $\left(x - a\right)$ must be a factor of the expression.

Therefore, we have three factors for the given expression:
$\textcolor{w h i t e}{\text{XXX}} \left(x - \left(- 1\right)\right) \left(x - \left(- 2\right)\right) \left(x - \left(- 3\right)\right)$
$\textcolor{w h i t e}{\text{XXXXsXXXXXXXXXXXXXXXX}} = \left(x + 1\right) \left(x + 2\right) \left(x + 3\right)$

Since the given expression is of degree $3$, it has at most $3$ factors and we have found them all.

Aug 7, 2017

${x}^{3} + 6 {x}^{2} + 11 x + 6 = \left(x + 1\right) \left(x + 2\right) \left(x + 3\right)$

#### Explanation:

Given:

${x}^{3} + 6 {x}^{2} + 11 x + 6$

You are actually likely to encounter this pattern of coefficients $1 , 6 , 11 , 6$ often.

For illustration purposes, let us see what happens if we start by applying a standard method used when solving a general cubic:

$\textcolor{w h i t e}{}$
Tschirnhaus transformation

In order to simplify any cubic of the form $a {x}^{3} + b {x}^{2} + c x + d$ we can apply a linear substitution $t = x + \frac{b}{3 a}$ to give a cubic in the form $a {t}^{3} + e t + f$ with no quadratic term. This is the simplest form of Tschirnhaus transformation.

In our example, $a = 1$ and $b = 6$, so we want $t = x + 2$.

Note that:

${\left(x + 2\right)}^{3} = {x}^{3} + 6 {x}^{2} + 12 x + 8$

So we find:

${x}^{3} + 6 {x}^{2} + 11 x + 6 = {\left(x + 2\right)}^{3} - \left(x + 2\right)$

$\textcolor{w h i t e}{{x}^{3} + 6 {x}^{2} + 11 x + 6} = {t}^{3} - t$

$\textcolor{w h i t e}{{x}^{3} + 6 {x}^{2} + 11 x + 6} = t \left({t}^{2} - 1\right)$

$\textcolor{w h i t e}{{x}^{3} + 6 {x}^{2} + 11 x + 6} = t \left(t - 1\right) \left(t + 1\right)$

$\textcolor{w h i t e}{{x}^{3} + 6 {x}^{2} + 11 x + 6} = \left(x + 2\right) \left(x + 1\right) \left(x + 3\right)$

In more sensible order:

${x}^{3} + 6 {x}^{2} + 11 x + 6 = \left(x + 1\right) \left(x + 2\right) \left(x + 3\right)$

Sep 17, 2017

${x}^{3} + 6 {x}^{2} + 11 x + 6 = \left(x + 1\right) \left(x + 2\right) \left(x + 3\right)$

#### Explanation:

It's probably worth mentioning that there's another way to solve this factoring problem if you have a calculator handy.

Given:

${x}^{3} + 6 {x}^{2} + 11 x + 6$

Since all of the coefficients of this polynomial are positive, we get an easy to calculate number if we put $x = 100$, namely:

$\textcolor{red}{1} {\left(\textcolor{b l u e}{100}\right)}^{3} + \textcolor{p u r p \le}{6} {\left(\textcolor{b l u e}{100}\right)}^{2} + \textcolor{b r o w n}{11} \left(\textcolor{b l u e}{100}\right) + \textcolor{g r e e n}{6} = \textcolor{red}{1} \textcolor{p u r p \le}{06} \textcolor{b r o w n}{11} \textcolor{g r e e n}{06}$

That is, the number formed by writing two digits for each coefficient (apart from the leading one, which does not need more than $1$ digit).

Then factors of the form $\left(x + k\right)$ correspond to integer factors with digits "$10 k$".

So we can try dividing $1061106$ by various numbers to see if we get whole results.

The rational root theorem can be used before that to let us know that the only possible rational zeros are:

$\pm 1 , \pm 2 , \pm 3 , \pm 6$

Hence we want to try factors: $101$, $102$, $103$, $106$.

In fact we find:

$\frac{1061106}{101} = 10506$

and:

$\frac{10506}{102} = 103$

Hence we can deduce that:

${x}^{3} + 6 {x}^{2} + 11 x + 6 = \left(x + 1\right) \left(x + 2\right) \left(x + 3\right)$