Answer:
The fourth term of the expansion is -220 * x^9 * y^3
Step-by-step explanation:
Question:
Find the fourth term in (x-y)^12
Solution:
Notation: "n choose k", or combination of k objects from n objects,
C(n,k) = n! / ( k! (n-k)! )
For example, C(12,4) = 12! / (4! 8!) = 495
Using the binomial expansion formula
(a+b)^n
= C(n,0)a^n + C(n,1)a^(n-1)b + C(n,2)a^(n-2)b^2 + C(n,3)a^(n-3)b^3 + C(n,4)a^(n-4)b^4 +....+C(n,n)b^n
For (x-y)^12, n=12, k=3, a=x, b=-y, and the fourth term is
C(n,3)a^(n-3)b^3
=C(12,3) * x^(12-3) * (-y)^(3)
= 220*x^9*(-y)^3
= -220 * x^9 * y^3
Answer:
j² - 5j²k - 2
Step-by-step explanation:
3j² - j²k - 6 - 4j²k - 2j² + 4
To simplify this polynomial, we can collect like terms. A term is number(s) or variable(s) that are grouped together by multiplication. <u>Like terms have the same variable and exponent</u>.
We have three groups of like terms:
The j-squares (j²), the j-squared k (j²k) and the constants (no variable).
Remember to include the negatives!
The j-squares are: 3j² ; -2j²
The j-squares k are: - j²k ; - 4j²k
The constants are: - 6 ; 4
Simplify:
3j² - j²k - 6 - 4j²k - 2j² + 4
Rearrange the polynomial by like terms
= (- j²k - 4j²k) + (3j² - 2j²) + (- 6 + 4)
Add or subtract the like terms
= (-5j²k) + (j²) + (-2)
Remove brackets and rearrange so the negative is not first
= j² + - 5j²k + - 2
Simplify where two signs are together. Adding a negative is subtraction.
= j² - 5j²k - 2 Simplified
