# How do you factor completely P(x)=x^3-6x^2+11x-6?

May 3, 2017

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

#### Explanation:

Given:

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

Note that linear factors correspond to zeros. That is, $x = a$ is a zero if and only if $\left(x - a\right)$ is a factor.

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Rational Roots Theorem

By the Rational Roots Theorem, any rational zeros of $P \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 6$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros of $P \left(x\right)$ are:

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

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Descartes' Rule of Signs

Note that the pattern of the signs of the coefficients of $P \left(x\right)$ is $+ - + -$. With $3$ changes, this means that $P \left(x\right)$ has $3$ or $1$ positive real zeros.

The pattern of the signs of the coefficients of $P \left(- x\right)$ is $- - - -$. With no changes, this means that $P \left(x\right)$ has no negative real zeros.

So, in combination with the Rational Roots Theorem, we can deduce that the only possible rational zeros of $P \left(x\right)$ are:

$1 , 2 , 3 , 6$

We could try each of these values in turn and find that:

$P \left(1\right) = P \left(2\right) = P \left(3\right) = 0$

Hence $P \left(x\right) = \left(x - 1\right) \left(x - 2\right) \left(x - 3\right)$

but here's another way...

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Sum of coefficients shortcut

Note that the sum of the coefficients of $P \left(x\right)$ is zero. That is:

$1 - 6 + 11 - 6 = 0$

Hence we can deduce that $x = 1$ is a zero of $P \left(x\right)$ and $\left(x - 1\right)$ a factor:

${x}^{3} - 6 {x}^{2} + 11 x - 6 = \left(x - 1\right) \left({x}^{2} - 5 x + 6\right)$

Then to factor the remaining quadratic note that $5 = 2 + 3$ and $6 = 2 \cdot 3$, and hence:

${x}^{2} - 5 x + 6 = \left(x - 2\right) \left(x - 3\right)$

Either way we arrive at our result:

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