(2x-y^2)
The expression (2x - y^2) is a polynomial.
How to save (2x-y^2)^9 by finding in the term of x^7y^5
To expand the expression (2x - y^2)^9 and find the term with x^7y^5, you can use the binomial theorem.
The binomial theorem states that for any positive integer n and any real numbers a and b, the expression (a + b)^n can be expanded as:
(a + b)^n = C(n, 0) * a^n * b^0 + C(n, 1) * a^(n-1) * b^1 + C(n, 2) * a^(n-2) * b^2 + ... + C(n, n-1) * a^1 * b^(n-1) + C(n, n) * a^0 * b^n
where C(n, k) represents the binomial coefficient, given by:
C(n, k) = n! / (k! * (n - k)!)
In this case, a = 2x and b = -y^2. We are looking for the term with x^7y^5, which corresponds to when k = 7 and (n - k) = 5.
So, substitute these values into the formula:
C(9, 7) * (2x)^7 * (-y^2)^5
C(9, 7) = 9! / (7! * (9 - 7)!) = 36
(2x)^7 = (2^7)(x^7) = 128x^7
(-y^2)^5 = (-1)^5 * (y^2)^5 = -y^10
Therefore, the term with x^7y^5 in the expansion of (2x - y^2)^9 is:
36 * 128x^7 * (-y^10) = -4608x^7y^10
Find the term independent of x in (4x^3-2÷x^2)^5
To find the term independent of x in the expansion of (4x^3 - 2/x^2)^5, we need to determine the term in which all the x's cancel out.
Using the binomial theorem, the general term in the expansion of (a + b)^n is given by:
C(n, k) * a^(n-k) * b^k
In this case, a = 4x^3 and b = -2/x^2.
For the term to be independent of x, the exponents of x in both a and b must add up to zero. In other words, we need n - k = 0.
So, substitute n - k = 0 into the formula:
C(5, k) * (4x^3)^(0) * (-2/x^2)^k
C(5, k) = 5! / (k! * (5 - k)!)
(4x^3)^0 = 1
(-2/x^2)^k = (-2^k)/(x^(2k))
Since (4x^3)^0 is always 1 regardless of k, we can ignore it in our calculation.
Now, let's determine which value of k will make the term independent of x, or in other words, cancel all the x's.
To make the x's cancel out, we need the exponent of x in (-2/x^2)^k to be equal to zero. So 2k = 0.
Solving for k, we find that k = 0.
Substituting k = 0 back into the formula:
Term independent of x = C(5, 0) * (4x^3)^(5 - 0) * (-2/x^2)^0
C(5, 0) = 5! / (0! * (5 - 0)!) = 1
(4x^3)^(5 - 0) = (4x^3)^5 = 1024x^15
(-2/x^2)^0 = 1
Therefore, the term independent of x in the expansion of (4x^3 - 2/x^2)^5 is:
1 * 1024x^15 * 1 = 1024x^15
Find expression (x+3)^4 by using binomial theory by combination to find the missing factor for any positive integer
To find the expression (x + 3)^4 using the binomial theorem, we can use the formula:
(a + b)^n = C(n, 0) * a^n * b^0 + C(n, 1) * a^(n-1) * b^1 + C(n, 2) * a^(n-2) * b^2 + ... + C(n, n-1) * a^1 * b^(n-1) + C(n, n) * a^0 * b^n
In this case, a = x and b = 3, and we want to expand it to the 4th power.
So substituting the values into the formula:
(x + 3)^4 = C(4, 0) * x^4 * 3^0 + C(4, 1) * x^3 * 3^1 + C(4, 2) * x^2 * 3^2 + C(4, 3) * x^1 * 3^3 + C(4, 4) * x^0 * 3^4
Now, let's calculate the values of the binomial coefficients C(n, k) for the different values of k:
C(4, 0) = 4! / (0! * (4 - 0)!) = 1
C(4, 1) = 4! / (1! * (4 - 1)!) = 4
C(4, 2) = 4! / (2! * (4 - 2)!) = 6
C(4, 3) = 4! / (3! * (4 - 3)!) = 4
C(4, 4) = 4! / (4! * (4 - 4)!) = 1
Plugging in these values:
(x + 3)^4 = 1 * x^4 * 3^0 + 4 * x^3 * 3^1 + 6 * x^2 * 3^2 + 4 * x^1 * 3^3 + 1 * x^0 * 3^4
Simplifying the expression:
(x + 3)^4 = x^4 + 12x^3 + 54x^2 + 108x + 81
Given the expression (2÷x^2-x)^15
To simplify the expression (2/x^2 - x)^15, we can apply the binomial theorem.
The binomial theorem states that for any positive integer n and any real numbers a and b, the expression (a + b)^n can be expanded as:
(a + b)^n = C(n, 0) * a^n * b^0 + C(n, 1) * a^(n-1) * b^1 + C(n, 2) * a^(n-2) * b^2 + ... + C(n, n-1) * a^1 * b^(n-1) + C(n, n) * a^0 * b^n
In this case, a = 2/x^2 and b = -x, and we are expanding it to the 15th power.
So substituting the values into the formula:
(2/x^2 - x)^15 = C(15, 0) * (2/x^2)^15 * (-x)^0 + C(15, 1) * (2/x^2)^(15-1) * (-x)^1 + C(15, 2) * (2/x^2)^(15-2) * (-x)^2 + ... + C(15, 14) * (2/x^2)^1 * (-x)^(15-1) + C(15, 15) * (2/x^2)^0 * (-x)^15
Now, let's calculate the values of the binomial coefficients C(n, k) for the different values of k:
C(15, 0) = 15! / (0! * (15 - 0)!) = 1
C(15, 1) = 15! / (1! * (15 - 1)!) = 15
C(15, 2) = 15! / (2! * (15 - 2)!) = 105
C(15, 3) = 15! / (3! * (15 - 3)!) = 455
C(15, 4) = 15! / (4! * (15 - 4)!) = 1365
C(15, 5) = 15! / (5! * (15 - 5)!) = 3003
C(15, 6) = 15! / (6! * (15 - 6)!) = 5005
C(15, 7) = 15! / (7! * (15 - 7)!) = 6435
C(15, 8) = 15! / (8! * (15 - 8)!) = 6435
C(15, 9) = 15! / (9! * (15 - 9)!) = 5005
C(15, 10) = 15! / (10! * (15 - 10)!) = 3003
C(15, 11) = 15! / (11! * (15 - 11)!) = 1365
C(15, 12) = 15! / (12! * (15 - 12)!) = 455
C(15, 13) = 15! / (13! * (15 - 13)!) = 105
C(15, 14) = 15! / (14! * (15 - 14)!) = 15
C(15, 15) = 15! / (15! * (15 - 15)!) = 1
Now, let's simplify the expression:
(2/x^2 - x)^15 = 1 * (2/x^2)^15 * (-x)^0 + 15 * (2/x^2)^14 * (-x)^1 + 105 * (2/x^2)^13 * (-x)^2 + ... + 15 * (2/x^2)^1 * (-x)^14 + 1 * (2/x^2)^0 * (-x)^15
Simplifying further:
(2/x^2 - x)^15 = (2^15 * x^15) / (x^30) + 15 * (2^14 * x^14) / (x^28) - 105 * (2^13 * x^13) / (x^26) + ... + 15 * (2 * x) / (x^2) - x^15
Simplifying the exponents and coefficients:
(2/x^2 - x)^15 = 2^15 * x^(15-30) + 15 * 2^14 * x^(14-28) - 105 * 2^13 * x^(13-26) + ... + 15 * 2 * x^(1-2) - x^15
(2/x^2 - x)^15 = 32768/x^15 + 16384/x^14 - 5460/x^13 + ... + 30/x - x^15
The first one is a trick question.
The expansion of (2x-y^2)^9 has all even powers of y.
So there is no term with x^7y^5.