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

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Answer 1 · 2016-05-15T12:09:29

$y = - x^{5} + 5 x^{4} - 10 x^{3} + 10 x^{2} - x + 1$ $y = -x^5+5x^4-10x^3+10x^2-x+1$

Notice that ${(x - 1)}^{2} {(x - 1)}^{3} = {(x - 1)}^{5}$ $(x-1)^2(x-1)^3 = (x-1)^5$

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.

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