# How do you determine the binomial factors of x^3-x^2-49x+49?

Dec 4, 2016

The binomial factors of ${x}^{3} - {x}^{2} - 49 x + 49$ are:

$\left({x}^{2} - 49\right)$, $\left(x - 1\right)$, $\left(x - 7\right)$, $\left(x + 7\right)$

#### Explanation:

The given cubic factors by grouping:

${x}^{3} - {x}^{2} - 49 x + 49 = \left({x}^{3} - {x}^{2}\right) - \left(49 x - 49\right)$

$\textcolor{w h i t e}{{x}^{3} - {x}^{2} - 49 x + 49} = {x}^{2} \left(x - 1\right) - 49 \left(x - 1\right)$

$\textcolor{w h i t e}{{x}^{3} - {x}^{2} - 49 x + 49} = \left({x}^{2} - 49\right) \left(x - 1\right)$

Note that both $\textcolor{b l u e}{\left({x}^{2} - 49\right)}$ and $\textcolor{b l u e}{\left(x - 1\right)}$ are binomial factors.

The difference of squares identity can be written:

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

Hence we find:

${x}^{2} - 49 = {x}^{2} - {7}^{2} = \left(x - 7\right) \left(x + 7\right)$

So $\textcolor{b l u e}{\left(x - 7\right)}$ and $\textcolor{b l u e}{\left(x + 7\right)}$ are also binomial factors.

Multiplying either of these by $\left(x - 1\right)$ will result in a trinomial, so there are no other binomial factors:

$\left(x - 7\right) \left(x - 1\right) = {x}^{2} - 8 x + 7$

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

The complete list of polynomial factors of ${x}^{3} - {x}^{2} - 49 x + 49$ in descending degree is:

${x}^{3} - {x}^{2} - 49 x + 49$

${x}^{2} - 49$

${x}^{2} - 8 x + 7$

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

$x - 7$

$x + 7$

$x - 1$

$1$