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

Oct 15, 2016

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

#### Explanation:

$\textcolor{b l u e}{1} {x}^{3} - {x}^{2} - 7 x + \textcolor{red}{3}$

To factor this expression, first list all the possible rational zeros.

The possible rational zeros are the factors of the constant $\textcolor{red}{3}$ divided by the factors of the leading coefficient $\textcolor{b l u e}{1}$.

$x = \frac{\pm 3}{\pm 1} = \pm 1 , \pm 3$

Next, pick one of the possible rational zeros and test it with synthetic division. If the remainder is zero, it is a zero of the equation. If the possible rational zero is c, the factor is $\left(x - c\right)$.

I will test $x = 3$ by synthetic division, which represents $\left({x}^{3} - {x}^{2} - 7 x + 3\right) \div \left(x - 3\right)$. If the remainder is zero, the factor is $\left(x - 3\right)$.

$3 | 1 \textcolor{w h i t e}{a} - 1 \textcolor{w h i t e}{a} - 7 \textcolor{w h i t e}{\textcolor{w h i t e}{a}} - 3$
$\textcolor{w h i t e}{a a} \downarrow \textcolor{w h i t e}{a a a} 3 \textcolor{w h i t e}{a {a}^{1}} 6 \textcolor{w h i t e}{{a}^{1}} - 3$
$\textcolor{w h i t e}{{a}^{1} a} \textcolor{m a \ge n t a}{1} \textcolor{w h i t e}{a {a}^{1} a} \textcolor{m a \ge n t a}{2} \textcolor{w h i t e}{a} \textcolor{m a \ge n t a}{- 1} \textcolor{w h i t e}{a a a a} \textcolor{\mathmr{and} a n \ge}{0}$

The remainder is $\textcolor{\mathmr{and} a n \ge}{z e r o}$, which means $x = 3$ is a zero.

The results of synthetic division are the coefficients of the polynomial quotient of the division of polynomials listed above.

$\textcolor{m a \ge n t a}{1} {x}^{2} + \textcolor{m a \ge n t a}{2} x \textcolor{m a \ge n t a}{- 1}$

This polynomial is not factorable. So ${x}^{3} - {x}^{2} - 7 x + 3$ factors to
$\left(x - 3\right) \left({x}^{2} + 2 x - 1\right)$.

Oct 16, 2016

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

$\textcolor{w h i t e}{{x}^{3} - {x}^{2} - 7 x + 3} = \left(x - 3\right) \left(x + 1 - \sqrt{2}\right) \left(x + 1 + \sqrt{2}\right)$

#### Explanation:

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

By the rational roots theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $3$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1 , \pm 3$

We find:

$f \left(3\right) = 27 - 9 - 21 + 3 = 0$

So $x = 3$ is a zero and $\left(x - 3\right)$ a factor:

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

The remaining quadratic can be factored using irrational coefficients by completing the square and using the difference of squares identity:

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

with $a = \left(x + 1\right)$ and $b = \sqrt{2}$ as follows:

${x}^{2} + 2 x - 1 = {x}^{2} + 2 x + 1 - 2$

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

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

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

Putting it all together:

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