Use the Binomial Theorem to expand the binomial.
left parenthesis s plus 3 v right parenthesis Superscript 5 baseline
The Binomial Theorem states that for any positive integer n, the expansion of the binomial (a + b)^n can be given by the following 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, the binomial is (s + 3v)^5.
By substituting a = s and b = 3v, we can apply the Binomial Theorem to expand the binomial:
(s + 3v)^5 = C(5, 0) * s^5 * (3v)^0 + C(5, 1) * s^4 * (3v)^1 + C(5, 2) * s^3 * (3v)^2 + C(5, 3) * s^2 * (3v)^3 + C(5, 4) * s^1 * (3v)^4 + C(5, 5) * s^0 * (3v)^5
Now, let's simplify each term:
= 1 * s^5 * 1 + 5 * s^4 * 3v + 10 * s^3 * (3v)^2 + 10 * s^2 * (3v)^3 + 5 * s^1 * (3v)^4 + 1 * s^0 * (3v)^5
= s^5 + 15s^4v + 90s^3v^2 + 270s^2v^3 + 405sv^4 + 243v^5
Therefore, the binomial (s + 3v)^5 expands to s^5 + 15s^4v + 90s^3v^2 + 270s^2v^3 + 405sv^4 + 243v^5.
To expand the binomial (s + 3v)^5 using the Binomial Theorem, you need to 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
Where:
- (a + b) is the binomial you want to expand.
- n is the power to which the binomial is raised.
- C(n, k) represents the binomial coefficient, which is the number of ways to choose k items from a set of n items.
Applying this formula to the given binomial (s + 3v)^5, we have:
n = 5
a = s
b = 3v
Using the formula, the expansion of (s + 3v)^5 will be:
C(5, 0) * s^5 * (3v)^0 + C(5, 1) * s^4 * (3v)^1 + C(5, 2) * s^3 * (3v)^2 + C(5, 3) * s^2 * (3v)^3 + C(5, 4) * s^1 * (3v)^4 + C(5, 5) * s^0 * (3v)^5
Now let's calculate each term step by step:
Term 1: C(5, 0) * s^5 * (3v)^0
Since C(5, 0) = 1 and (3v)^0 = 1, this term simplifies to s^5.
Term 2: C(5, 1) * s^4 * (3v)^1
C(5, 1) = 5, so this term becomes 5s^4 * 3v.
Term 3: C(5, 2) * s^3 * (3v)^2
C(5, 2) = 10, so this term becomes 10s^3 * (3v)^2.
Term 4: C(5, 3) * s^2 * (3v)^3
C(5, 3) = 10, so this term becomes 10s^2 * (3v)^3.
Term 5: C(5, 4) * s^1 * (3v)^4
C(5, 4) = 5, so this term becomes 5s^1 * (3v)^4.
Term 6: C(5, 5) * s^0 * (3v)^5
Since C(5, 5) = 1 and s^0 = 1, this term simplifies to (3v)^5.
Finally, combining all the terms, the expanded form of (s + 3v)^5 is:
s^5 + 5s^4 * 3v + 10s^3 * (3v)^2 + 10s^2 * (3v)^3 + 5s^1 * (3v)^4 + (3v)^5
To expand the binomial expression (s + 3v)^5 using the Binomial Theorem, we can use the formula:
(x + y)^n = C(n, 0) * x^(n-0) * y^0 + C(n, 1) * x^(n-1) * y^1 + C(n, 2) * x^(n-2) * y^2 + ... + C(n, r) * x^(n-r) * y^r + ... + C(n, n) * x^0 * y^n
where:
- C(n, r) represents the binomial coefficient (n choose r),
- x is the first term of the binomial expression,
- y is the second term of the binomial expression, and
- n is the power to which the binomial expression is raised.
In this case, we have:
x = s
y = 3v
n = 5
Using the formula, we can expand the binomial as follows:
(s + 3v)^5 = C(5, 0) * s^(5-0) * (3v)^0 + C(5, 1) * s^(5-1) * (3v)^1 + C(5, 2) * s^(5-2) * (3v)^2 + C(5, 3) * s^(5-3) * (3v)^3 + C(5, 4) * s^(5-4) * (3v)^4 + C(5, 5) * s^0 * (3v)^5
Expanding each term using the binomial coefficients, we get:
(s + 3v)^5 = 1 * s^5 * 3v^0 + 5 * s^4 * 3v^1 + 10 * s^3 * (3v)^2 + 10 * s^2 * (3v)^3 + 5 * s^1 * (3v)^4 + 1 * s^0 * (3v)^5
Simplifying each term by raising the variables to the respective powers:
(s + 3v)^5 = s^5 + 15s^4v + 90s^3v^2 + 270s^2v^3 + 405sv^4 + 243v^5
So, the binomial expansion of (s + 3v)^5 using the Binomial Theorem is: s^5 + 15s^4v + 90s^3v^2 + 270s^2v^3 + 405sv^4 + 243v^5.