Pascal's Triangle is structured as follows:
Row 0: 1
Row 1: 1 1
Row 2: 1 2 1
Row 3: 1 3 3 1
Row 4: 1 4 6 4 1
Row 5: 1 5 10 10 5 1
To find \((3x - 2y)^5\), we can use the Binomial Theorem, which states:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
For our case, we let \(a = 3x\) and \(b = -2y\), and \(n = 5\).
Using the coefficients from Row 5 of Pascal's Triangle (which correspond to the binomial coefficients \(\binom{5}{k}\)), we have:
- \(\binom{5}{0} = 1\)
- \(\binom{5}{1} = 5\)
- \(\binom{5}{2} = 10\)
- \(\binom{5}{3} = 10\)
- \(\binom{5}{4} = 5\)
- \(\binom{5}{5} = 1\)
Now we can expand \((3x - 2y)^5\):
\[ (3x - 2y)^5 = \sum_{k=0}^{5} \binom{5}{k} (3x)^{5-k} (-2y)^k \]
Calculating each term:
-
For \(k=0\): \[ \binom{5}{0} (3x)^5 (-2y)^0 = 1 \cdot (3x)^5 \cdot 1 = 243x^5 \]
-
For \(k=1\): \[ \binom{5}{1} (3x)^4 (-2y)^1 = 5 \cdot (81x^4) \cdot (-2y) = -810x^4y \]
-
For \(k=2\): \[ \binom{5}{2} (3x)^3 (-2y)^2 = 10 \cdot (27x^3) \cdot (4y^2) = 1080x^3y^2 \]
-
For \(k=3\): \[ \binom{5}{3} (3x)^2 (-2y)^3 = 10 \cdot (9x^2) \cdot (-8y^3) = -720x^2y^3 \]
-
For \(k=4\): \[ \binom{5}{4} (3x)^1 (-2y)^4 = 5 \cdot (3x) \cdot (16y^4) = 240xy^4 \]
-
For \(k=5\): \[ \binom{5}{5} (3x)^0 (-2y)^5 = 1 \cdot 1 \cdot (-32y^5) = -32y^5 \]
Putting it all together:
\[ (3x - 2y)^5 = 243x^5 - 810x^4y + 1080x^3y^2 - 720x^2y^3 + 240xy^4 - 32y^5 \]
Thus, the expansion of \((3x - 2y)^5\) is:
\[ \boxed{243x^5 - 810x^4y + 1080x^3y^2 - 720x^2y^3 + 240xy^4 - 32y^5} \]