# How do you factor the expression x^4+4x^3+6x^2+4x+1?

Feb 3, 2016

${x}^{4} + 4 {x}^{3} + 6 {x}^{2} + 4 x + 1 = {\left(x + 1\right)}^{4}$

#### Explanation:

The coefficients of this quartic are $1 , 4 , 6 , 4 , 1$ - the fifth row of Pascal's triangle.

These are the coefficients of the binomial expansion of ${\left(a + b\right)}^{4}$.

In our example, $a = x$ and $b = 1$

Alternatively, first notice that if we give the coefficients alternating signs then they sum to $0$:

$1 - 4 + 6 - 4 + 1 = 0$

Hence $x = - 1$ is a zero of ${x}^{4} + 4 {x}^{3} + 6 {x}^{2} + 4 x + 1$ and $\left(x + 1\right)$ is a factor...

${x}^{4} + 4 {x}^{3} + 6 {x}^{2} + 4 x + 1 = \left(x + 1\right) \left({x}^{3} + 3 {x}^{2} + 3 x + 1\right)$

If we give the coefficients of the remaining cubic factor alternating signs then again they add to $0$:

$1 - 3 + 3 - 1 = 0$

Hence $x = - 1$ is a zero of ${x}^{3} + 3 {x}^{2} + 3 x + 1$ and $\left(x + 1\right)$ is a factor...

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

Similarly we find ${x}^{2} + 2 x + 1 = \left(x + 1\right) \left(x + 1\right)$ and we have found all of the linear factors.