Expanding Binomial with Coefficients of x² and x³
The image shows several mathematical problems, but I will focus on question number 2, as requested:
"2. The coefficient of x^2 and x^3 in the expansion of (3-2x)^5 are a and b respectively. Find the value of p = a-b."
To find the coefficients of x² and x³ in the expansion of (3-2x)⁵, we can use the binomial theorem.
The binomial theorem states that
(a + b)ⁿ = Σ from k = 0 to n (n choose k) * a^(n-k) * b^k.
For (3 - 2x)⁵, we have a = 3 and b = -2x. Now we need to find the terms where the exponent of x is 2 and 3.
The general term in binomial expansion is T(k+1) = (n choose k) * a^(n-k) * b^k.
For the x² term (k = 2):
T(3) = (5 choose 2) * 3^(5-2) * (-2x)²
T(3) = 10 * 3³ * 4x²
T(3) = 10 * 27 * 4x²
T(3) = 1080 x²
So, a = 1080.
For the x³ term (k = 3):
T(4) = (5 choose 3) * 3^(5-3) * (-2x)³
T(4) = 10 * 3² * -8x³
T(4) = 10 * 9 * -8x³
T(4) = -720 x³
So, b = -720.
Finally, we have to find p = a - b:
p = a - b
p = 1080 - (-720)
p = 1080 + 720
p = 1800.
Therefore, the value of p is 1800.
