# How do you determine the binomial factors of x^3+3x^2+3x+1?

May 28, 2017

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

#### Explanation:

Using the rational root theorem gives:

$p = 1 , q = 1$

All values of $\pm \frac{p}{q}$ are $- 1$ and $1$.

Now, plug in each value to the polynomial. By the remainder theorem, if the output is zero, the input must be a root.

${\left(- 1\right)}^{3} + 3 {\left(- 1\right)}^{2} + 3 \left(- 1\right) + 1 = - 1 + 3 - 3 + 1 = 0$
$\therefore$ there is a root at $x = - 1$

${\left(1\right)}^{3} + 3 {\left(1\right)}^{2} + 3 \left(1\right) + 1 = 1 + 3 + 3 + 1 = 8$
$\therefore$ there is NOT a root at $x = 1$

Factors of the polynomial: $\left(x - \text{root}\right) = \left(x - \left(- 1\right)\right) = \left(x + 1\right)$
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Now, divide the polynomial by $\left(x + 1\right)$:

$\frac{{x}^{3} + 3 {x}^{2} + 3 x + 1}{x + 1} = \frac{{x}^{3} + {x}^{2} + {x}^{2} + x + {x}^{2} + x + x + 1}{x + 1}$

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

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

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

You could do the same process with this new polynomial, or you could recognize it as a perfect square.

Either way, factoring this polynomial results in ${\left(x + 1\right)}^{2}$

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So the final answer is $\left(x + 1\right) {\left(x + 1\right)}^{2} = {\left(x + 1\right)}^{3}$