factor (1-y)^8
To factor the expression (1-y)^8, we can use the binomial theorem. The binomial theorem states that for any real numbers a and b and any positive integer n, the expansion of (a+b)^n can be written as the sum of the binomial coefficients multiplied by the respective powers of a and b.
In this case, a is 1 and b is -y. So, the expression (1-y)^8 can be expanded using the binomial theorem as:
(1-y)^8 = C(8, 0) * 1^8 * (-y)^0 + C(8, 1) * 1^7 * (-y)^1 + C(8, 2) * 1^6 * (-y)^2 + C(8, 3) * 1^5 * (-y)^3 + C(8, 4) * 1^4 * (-y)^4 + C(8, 5) * 1^3 * (-y)^5 + C(8, 6) * 1^2 * (-y)^6 + C(8, 7) * 1^1 * (-y)^7 + C(8, 8) * 1^0 * (-y)^8
In this expansion, C(n, k) represents the binomial coefficient, which can be calculated as C(n, k) = n! / (k! * (n-k)!), where n! is the factorial of n.
Now, let's simplify this expression:
C(8, 0) * 1^8 * (-y)^0 = 1 * 1 * (-y)^0 = 1
C(8, 1) * 1^7 * (-y)^1 = 8 * 1 * (-y)^1 = -8y
C(8, 2) * 1^6 * (-y)^2 = 28 * 1 * (-y)^2 = 28y^2
C(8, 3) * 1^5 * (-y)^3 = 56 * 1 * (-y)^3 = -56y^3
C(8, 4) * 1^4 * (-y)^4 = 70 * 1 * (-y)^4 = 70y^4
C(8, 5) * 1^3 * (-y)^5 = 56 * 1 * (-y)^5 = -56y^5
C(8, 6) * 1^2 * (-y)^6 = 28 * 1 * (-y)^6 = 28y^6
C(8, 7) * 1^1 * (-y)^7 = 8 * 1 * (-y)^7 = -8y^7
C(8, 8) * 1^0 * (-y)^8 = 1 * 1 * (-y)^8 = y^8
Putting it all together, we have:
(1-y)^8 = 1 - 8y + 28y^2 - 56y^3 + 70y^4 - 56y^5 + 28y^6 - 8y^7 + y^8