I am refreshing before taking Calc1 and there is a sequence my instructor gave us and I am struggling to follow the fractional exponent moves he's made. The problem is this: \begin{align*} 5x^2(2x-3)^{1/3}&(3x+2)^{1/2}+8x(2x-3)^{-2/3}(3x+2)^{3/2}\\ & = x(2x-3)^{-2/3}(3x+2)^{1/2}[5x(2x-3)+8(3x+2)]\\ & = x(2x-3)^{-2/3}(3x+2)^{1/2}[10x^2-15x+24x+16]\\ & = x(2x-3)^{-2/3}(3x+2)^{1/2}(10x^2+9x+16) \end{align*}
A) I don't remember ever being taught fractional exponents let alone taking the smallest exponent from all common terms (what does that even mean)
B) From the first format to the first equal sign, why was the problem essentially switched with the $5x$ etc that started the problem moved to the back, and then the original $8x(2x-3)^{-2/3}(3x+2)^{3/2}$ switches to the front and we lose the $8$ at the beginning and the last exponent turns from a $3/2$ to a $1/2$. It's one of those notions that's on the tip of my brain but I'm still not making that connection.
The greatest common divisor ($\gcd$) or greatest common factor (gcf) or highest common factor (hcf) of two integers is the largest number which is a divisor or factor of those integers.
The most efficient way to find the greatest common divisor is the Euclidean algorithm. For small integers, we can simply compare lists of divisors/factors to find the $\gcd$. For instance, suppose we wish to find the $\gcd$ of $60$ and $96$. The divisors/factors of $60$ are $1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60$. The divisors/factors of $96$ are $1, 2, 3, 4, 6, 8, 12, 24, 32, 48, 96$. Comparing these lists, we see by inspection that $\gcd(60, 96) = 12$.
If we factor both $60$ and $96$ into primes, we obtain \begin{align*} 60 & = 2^2 \cdot 3 \cdot 5\\ 96 & = 2^5 \cdot 3 \end{align*} Notice that $$\gcd(60, 96) = \gcd(2^2 \cdot 3^1 \cdot 5^1, 2^5 \cdot 3^1 \cdot 5^0) = 2^2 \cdot 3^1 \cdot 5^0 = 4 \cdot 3 \cdot 1 = 12$$ since $2^2$ is the highest power of $2$ which appears in the prime factorizations of both $60$ and $96$, $3^1$ is the highest power of $3$ which appears in the prime factorizations of $60$ and $96$, and $5^0$ is the highest power of $5$ which appears in the prime factorizations of both $60$ and $96$. Observe that the greatest common divisor can be found by taking the smaller exponent for each prime that appears in the factorizations of at least one of the numbers, and then taking the product of the resulting prime powers.
We can treat variables like prime factors. Hence, $$\gcd(x^3y^2, x^2y^4) = x^2y^2$$ since $x^2$ is the highest power of $x$ which is a factor of both $x^3$ and $x^2$ and $y^2$ is the highest power of $y$ which is a factor of both $y^2$ and $y^4$. Notice that in each case, we are taking the smaller exponent for each variable, then multiplying the resulting powers.
You were asked to factor the expression $$5x^2(2x-3)^{1/3}(3x+2)^{1/2}+8x(2x-3)^{-2/3}(3x+2)^{3/2}$$ Observe that \begin{align*} \gcd(5, 8) & = 1\\ \gcd(x^2, x) & = x\\ \gcd((2x - 3)^{1/3}, (2x - 3)^{-2/3}) & = (2x - 3)^{-2/3}\\ \gcd((3x + 2)^{1/2}, (3x + 2)^{3/2}) & = (3x + 2)^{1/2} \end{align*} where in each case we take the smaller exponent for each common factor. Hence, $$\gcd(5x^2(2x - 3)^{1/3}(3x + 2)^{1/2} , 8x(2x - 3)^{-2/3}(3x + 2)^{3/2}) = x(2x - 3)^{-2/3}(3x + 2)^{1/2}$$
Recall the following rules of exponents: $x^{r + s} = x^rx^s$ and $(x^r)^s = x^{rs}$. For instance, $$8^48^5 = \underbrace{(8 \cdot 8 \cdot 8 \cdot 8)}_{4~\text{factors}}\underbrace{(8 \cdot 8 \cdot 8 \cdot 8 \cdot 8)}_{5~\text{factors}} = \underbrace{8 \cdot 8 \cdot 8 \cdot 8 \cdot 8 \cdot 8 \cdot 8 \cdot 8 \cdot 8}_{9~\text{factors}} = 8^9$$ and $$(7^3)^2 = (\underbrace{7 \cdot 7 \cdot 7}_{3~\text{factors}})(\underbrace{7 \cdot 7 \cdot 7}_{3~\text{factors}}) = \underbrace{7 \cdot 7 \cdot 7 \cdot 7 \cdot 7 \cdot 7}_{6~\text{factors}} = 7^6$$
By definition, $x^{1/n} = \sqrt[n]{x}$ and $x^{m/n} = (x^{1/n})^m$. Thus, $$(x^{1/3})^3 = x$$ and $$x^{3/2} = (x^{1/2})^3 = (x^{1/2})(x^{1/2})(x^{1/2}) = x^{1/2 + 1/2 + 1/2}$$ Notice that our definitions are consistent with the rules of exponents we stated above.
Thus, we can factor $(3x + 2)^{3/2}$ as $$(3x + 2)^{3/2} = (3x + 2)^1(3x + 2)^{1/2} = (3x + 2)(3x + 2)^{1/2}$$ and $$(2x - 3)^{1/3} = (2x - 3)^1(2x - 3)^{-2/3} = (2x - 3)(2x - 3)^{-2/3}$$
Therefore, extracting the $\gcd$ from each term yields \begin{align*} & 5x^2(2x-3)^{1/3}(3x+2)^{1/2}+8x(2x-3)^{-2/3}(3x+2)^{3/2}\\ & \qquad = x(2x - 3)^{-2/3}(3x + 2)^{1/2}[5x(2x - 3) + 8(3x + 2)] \end{align*} I gather you understand the remaining steps of the calculation.