Which term in the expansion of ((1/2x^3)-(x^5))^8 is a constant?
THANKS!!
1
x is variable
3,5,8 is exponents
noo...it's asking after u expand the entire thing, which term is a constant ....for example
(2a-3)^2=
(2a-3)(2a-3)
4a^2-12a+9
Then the third term would be "constant" cause it doesn't have any variables attached to it...
To find the constant term in the expansion of ((1/2x^3)-(x^5))^8, we can use the Binomial Theorem. The Binomial Theorem states that for any two terms, (a + b)^n, the constant term can be found using the combination formula:
C(n, r) * a^(n-r) * b^r
Where C(n, r) represents the binomial coefficient, also known as "n choose r".
In our case, the expression ((1/2x^3)-(x^5))^8, the term (1/2x^3) is represented by 'a' and the term (-x^5) is represented by 'b'.
We want to find the constant term when we expand ((1/2x^3)-(x^5))^8, so we want to find the term where r = 8 - r, or equivalently, r = 4.
Plugging the values into the formula, we have:
C(8, 4) * (1/2x^3)^(8-4) * (-x^5)^4
C(8, 4) = 8! / (4! * (8-4)!) = 70
(1/2x^3)^(8-4) = (1/2x^3)^4 = (1/16x^12)
(-x^5)^4 = x^20
Putting everything together:
70 * (1/16x^12) * x^20 = 70/16x^(-12) * x^20 = 70/16x^(-12 + 20) = 70/16x^8 = 35/8x^8
Therefore, the term that represents the constant in the expansion of ((1/2x^3)-(x^5))^8 is 35/8x^8.