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

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Answer 1 · 2016-02-03T00:06:29

$x^4+4x^3+6x^2+4x+1 = (x+1)^4$

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

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These are the coefficients of the binomial expansion of $(a+b)^4$ .

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 $x^4+4x^3+6x^2+4x+1$ and $(x+1)$ is a factor...

$x^4+4x^3+6x^2+4x+1 = (x+1)(x^3+3x^2+3x+1)$

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 $x^3+3x^2+3x+1$ and $(x+1)$ is a factor...

$x^3+3x^2+3x+1 = (x+1)(x^2+2x+1)$

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

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