What is the standard form of #y-4x= -(x-1)^2(x-1)^3#?

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Answer 1 · 2016-05-15T12:09:29

#y = -x^5+5x^4-10x^3+10x^2-x+1#

Notice that #(x-1)^2(x-1)^3 = (x-1)^5#

The binomial theorem tells us that:

#(a+b)^n = sum_(k=0)^n ((n),(k)) a^(n-k)b^k#

where the binomial coefficient: #((n),(k)) = (n!)/(k! (n-k)!)#

The binomial coefficients can be found as rows in Pascal's triangle:

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The row #1, 5, 10, 10, 5, 1# gives us the binomial coefficients for #(a+b)^5#:

#(a+b)^5 = a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5#

Then put #a=x#, #b=-1# to find:

#(x-1)^5 = x^5-5x^4+10x^3-10x^2+5x-1#

So:

#y - 4x#

#= -(x-1)^2(x-1)^3#

#= -(x-1)^5#

#= -(x^5-5x^4+10x^3-10x^2+5x-1)#

#= -x^5+5x^4-10x^3+10x^2-5x+1#

Add #4x# to both ends to get:

#y = -x^5+5x^4-10x^3+10x^2-x+1#

This has the powers of #x# in descending order, which is the standard form.