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.