Efficient Factoring Part 1
Expressions With Denominators
In Calculus I, it pays if you can factor efficiently. For example, to find where a function $y = f(x)$ is increasing, take the derivative, $y = f'(x)$, and see where it is positive. When an expression is factored, it is written as products of factors. In most cases, it is simpler to determine where an expression is positive when it is it written as a product rather than a sum or difference. The sum of a positive and a negative value might be positive, or it might be negative. It might be zero. But the product of a positive and a negative value is always negative.
To factor out the common terms in a sum (or difference), factor the least power of each term.
Factor completely.
\[ x^2y^{5/3} + x^3yz \]The powers of $x$ are $2$ and $3$. The least of these is $2$, so factor out $x^2$. The powers of $y$ are $\frac{5}{3}$ and $1$. The least of these is $1$, so factor out $y^1$. There is no $z$ in the first term, so don't factor out any power of $z$.
\[ x^2y^{5/3} + x^3yz = x^2y \left(y^{2/3} + xz\right) \]I assume you already know how to do this, but take a closer look at the process. Factoring out $x^2$ means that each term is being divided by $x^2$. When $x^2$ is divided by $x^2$, the quotient is 1 and when $x^3$ is divided by $x^2$, the quotient is $x$. When dividing powers, subtract exponents. When $x^2$ is divided by $x^2$, the quotient is $x^{2-2} = x^0 = 1$, and when $x^3$ is divided by $x^2$, the quotient is $x^{3-2} = x^1 = x$.
When $y^{5/3}$ is divided by $y^1$, the quotient is $y^{5/3 - 1} = y^{2/3}$.
The algorithm is as follows: when factoring $x^a + x^b$, compare $a$ and $b$, and choose whichever is least. Factor out $x$ raised to this power. For example, if $a \leq b$, factor out $x^a$. This leaves $x^a(1 + x^{b-a})$. Because $a \leq b$, $b - a \geq 0$. This ensures there won't be any negative exponents; that is, there will be no terms with denominators.
The algorithm works with $z$ as well as $x$ and $y$. There is no $z$ in the first term. Think of this as $z^0$. The power of $z$ in the second term is $1$. The least of these is $0$, so factor out $z^0$. Because $z^0 = 1$, factor out $1$, which is the same as not factoring out anything.
This algorithm also works for expressions with denominators. Write the denominators as powers with negative exponents and use the least exponent rule described previously.
Example
Factor \[ \frac{x^2}{y^{5/3}} + \frac{yz}{x^3} \]Solution
Most students will find a common denominator first and then factor, but that is inefficient. It is faster and easier (after a little practice) to use the exact same algorithm you use when there are no denominators.
The first step is to write everything as a product of powers. Use negative exponents for the terms in the denominators.
\[ \frac{x^2}{y^{5/3}} + \frac{yz}{x^3} = x^2y^{-5/3} + yzx^{-3} \]Proceed exactly as before, comparing the exponents for each term. Factor out the power with the least exponent; when $x^a$ is factored out of $x^b$, the term that remains is $x^{b-a}$.
\begin{align*} \frac{x^2}{y^{5/3}} + \frac{yz}{x^3} & = x^2y^{-5/3} + yzx^{-3} \\ & = x^{-3}y^{-5/3} \left[\left(x^{2 - (-3)}y^{-5/3 - (-5/3)}\right) + \left(x^{-3 - (-3)}y^{1 - (-5/3)}z\right)\right] \\ & = x^{-3}y^{-5/3} \left(x^{5} + y^{7/3}z\right) \\ & = \frac{x^5 + y^{7/3}z}{x^3y^{5/3}} \end{align*}With a little practice, you should get to where you can do the subtraction step in your head.
Example
Factor \[ \frac{4x^3}{15\sqrt{y}} + \frac{8y^3}{3z^2} \]Solution
Write everything as a product of powers. Use negative exponents for the terms in the denominators.
\begin{align*} \frac{4x^3}{15\sqrt{y}} + \frac{8y^3}{3z^2} & = \frac{2^2 x^3}{(3 \cdot 5) \sqrt{y}} + \frac{2^3y^3}{3z^2} \\ & = \left(2^2 \cdot x^3 \cdot 3^{-1} \cdot 5^{-1} \cdot y^{-1/2}\right) + \left(2^3 \cdot y^3 \cdot 3^{-1} \cdot z^{-2}\right) \\ & = \left(2^2 \cdot 3^{-1} \cdot 5^{-1} \cdot x^3 \cdot y^{-1/2} \cdot z^{0}\right) + \left(2^3 \cdot 3^{-1} \cdot 5^{0} \cdot x^{0} \cdot y^3 \cdot z^{-2}\right) \\ \end{align*}I reordered the factors and included the zero powers to make it easier to compare the exponents.
The second step is to compare the exponents for each term. Factor out the power with the least exponent, and when you factor out $x^a$ from $x^b$, the term that remains is $x^{b-a}$.
\begin{align*} \lefteqn{\left(2^2 \cdot 3^{-1} \cdot 5^{-1} \cdot x^3 \cdot y^{-1/2} \cdot z^{0}\right) + \left(2^3 \cdot 3^{-1} \cdot 5^{0} \cdot z^{0} \cdot y^3 \cdot z^{-2}\right)} \\ & = 2^2 \cdot 3^{-1} \cdot 5^{-1} \cdot x^0 \cdot y^{-1/2} \cdot z^{-2} \cdot \left(2^{2-2} \cdot 3^{-1- (-1)} \cdot 5^{-1- (-1)} \cdot x^{3-3} \cdot y^{-1/2 - (-1/2)} \cdot z^{0-(-2)} \right. + \left.2^{3-2} \cdot 3^{-1 - (-1)} \cdot 5^{0 - (-1)} \cdot x^{3-0} \cdot y^{3 - (-1/2)} \cdot z^{-2 - (-2)}\right) \\ & = 2^2 \cdot 3^{-1} \cdot 5^{-1} \cdot y^{-1/2} \cdot z^{-2} \cdot \left(x^3 \cdot z^{2} + 2 \cdot 5^{1} \cdot y^{7/2)}\right) \\ & = \frac{4}{15\sqrt{y}z^2} \left(x^3z^{2} + 10y^{7/2}\right) \\ \end{align*}With a little practice, you should get to where you can do the subtractions in your head, which eliminates the second line in this calculation.
If you would like to try this, here are some you can practice on.
\[ \frac{2x}{e^x} - x^2 \] \[ \frac{e^x}{2\sqrt{x}} - \frac{x^2}{e^x} \] \[ \frac{e^x(x^2+1) + e^{x}(-2x)}{(x-1)^2} \] \[ x^2\sin 2x + \frac{\cos x}{\sqrt{x}} \]To see the answers, click here.