How do you use the Binomial Throrem to expand (x-2y)^5?

Answer 1

(x-2y)^5=x^5+(5C1)(x^4)(-2y)
+(5C2)(x^3)(-2y)^2
+(5C3)(x^2)(-2y)^3
+(5C4)(x)(-2y)^4+(-2y)^5
(a+b)^n=a^n+nC1*a^(n-1)b^1
+.....+b^n

Answer 2

Make a 3-column list. In column 1, row 5 from Pascal's Triangle (just 5C0 through 5C5): 1 5 10 10 5 1. In column 2, descending powers of x: x^5, x^4, x^3, x², x, and 1. In column 3, ascenting powers of (-2y): 1, -2y, 4y², -8y^3, 16y^4, -32y^5.

Then multiply across each row, for example, row 3 is 10 • x^3 • 4y² = 40x^3y², and that's the 3rd term of the expansion. Adding all the (unlike) terms you got by multiplying across gives you the entire expansion.

Answer 3

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

In your case a = x and b = 2y

x^5 - 5x^4*(2y) + 10x^3*(2y)^2 - 10x^2*(2y)^3 + 5x*(2y)^4 - (2y)^5
= x^5 - 10x^4y + 40x^3y^2 - 80x^2y^3 + 80xy^4 - 32y^5

Answer 4

(x-2y)^5 can be divided as ((x-2y)^2)*((x-2y)^3)
they are in the form of a-b whole sq. and a-b whole cube
(x^2-4xy+4y^2)*(x^3-6xy(x-2y)-8y^3)

Answer 5

The rth term for (a+b)^n = nCr a^(r) b^(n-r).
If b= -b then ie for (a-b)^n do suitable substitutions so instead of b^(n-r) u will have (-b)^(n-r). Hope that helps

Answer 6

5x-10y