Use Pascal's Triangle to expand $\left(3 y^2+z\right)^4$. Express your answer in simplest form.
To expand the binomial \((3y^2 + z)^4\), we use the Binomial Theorem. The theorem states that for any non-negative integer \(n\), the expansion of \((a + b)^n\) is given by:
\[(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\]
In this specific problem, we have \(a = 3y^2\), \(b = z\), and \(n = 4\). The coefficients \(\binom{n}{k}\) are the values from the 5th row of Pascal's Triangle, which are \(1, 4, 6, 4, 1\).
We substitute \(a = 3y^2\), \(b = z\), and \(n = 4\) into the binomial expansion formula:
\[(3y^2 + z)^4 = \binom{4}{0}(3y^2)^4(z)^0 + \binom{4}{1}(3y^2)^3(z)^1 + \binom{4}{2}(3y^2)^2(z)^2 + \binom{4}{3}(3y^2)^1(z)^3 + \binom{4}{4}(3y^2)^0(z)^4\]
Using the coefficients from Pascal's Triangle, this becomes:
\[(3y^2 + z)^4 = 1 \cdot (3y^2)^4 \cdot 1 + 4 \cdot (3y^2)^3 \cdot z + 6 \cdot (3y^2)^2 \cdot z^2 + 4 \cdot (3y^2)^1 \cdot z^3 + 1 \cdot 1 \cdot z^4\]
Now, we simplify each term by applying the power rules \((xy)^n = x^n y^n\) and \((x^m)^n = x^{mn}\):
Combining these, we get:
\[81y^8 + 108y^6z + 54y^4z^2 + 12y^2z^3 + z^4\]
The expansion of binomials is a fundamental technique in algebra and probability. For instance, in finance, if an investment has two components that fluctuate, one might use binomial expansions to model the overall growth of an account over several periods. By expanding expressions like \((1+r)^n\), analysts can predict total returns even when rates change or are applied to different parts of an asset. Understanding these patterns allows for more efficient calculations in various fields, from engineering physics to data science.