What is the binomial expansion of #(2x-1)^5#?

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Answer 1 · 2015-08-17T19:33:30

#color(red)((2x-1)^5 = 32x^5-80x^4+80x^3-40x^2+10x-1)#

Write out the sixth row of Pascal's triangle and make the appropriate substitutions.

Pascal's triangle is

The numbers in the fifth row are 1, 5, 10, 10, 5, 1.

They are the coefficients of the terms in a fifth order polynomial.

#(x+y)^5 = x^5 + 5x^4y + 10x^3y^2 + 10x^2y^3 + 5xy^4 + y^5#

But your polynomial is #(2x-1)^5#.

Let #w=2x# and #z=-1#.

Then your polynomial becomes

#(2x-1)^5 =(w+z)^5#.

#(w+z)^5= w^5 + 5w^4z + 10w^3z^2 + 10w^2z^3 + 5wz^4 + z^5#

Now re-insert the values of #w# and#z#

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

#(2x-1)^5 = 32x^5-80x^4+80x^3-40x^2+10x-1#