This section contains:
 Revisiting Factoring Quadratics
 Factoring Sum and Difference of Cubes
 Factoring and Solving with Polynomials
 Factoring and Solving with Exponents
 More Practice
Since factoring is so important in algebra, let’s revisit it first. Remember that we first learned factoring here in the Solving Quadratics by Factoring and Completing the Square section.
Revisiting Factoring Quadratics
Earlier, we learned how to “unfoil” a trinomial into two binomials:
\({{x}^{2}}11x60=\left( {x15} \right)\left( {x+4} \right)\)  Find two numbers that multiplied together are \(60\) and added together are \(11\). 
Sometimes we have to factor out some stuff before we do the foiling. We always want to do this first:
\(\begin{align}6{{x}^{3}}21{{x}^{2}}45x&=3x\left( {2{{x}^{2}}7x15} \right)\\&=3x\left( {2x+3} \right)\left( {x5} \right)\end{align}\)  We first have to take the greatest common factor (GCF) out, which is \(3x\). We then have “unfoil”, but don’t forget to bring the \(3x\) down as part of the answer. (Remember when factoring that the inner products and outer products have to add up to \(7x:\,\,3x+10x=7x\)). 
And don’t forget “grouping” when we have four terms (but it doesn’t always work – we’ll find other ways to solve in Graphing and Solving Polynomials later). Again, we are working with GCF’s to do this:
\(\displaystyle \begin{align}24xy+32{{x}^{2}}y+6z+8xz&=8xy\left( {3+4x} \right)+2z\left( {3+4x} \right)\\&=\left( {8xy+2z} \right)\left( {3+4x} \right)\end{align}\)  Separate the four terms into two sets of two terms. Take out the GCF out of each of the sets of terms. Then, because of the distributive property, take out the expression that’s repeated, use that as one factor, and use the coefficients of this expression as the other factor. 
You can factor a difference of squares, but not a sum of squares:
\(9{{x}^{2}}25=\left( {3x5} \right)\left( {3x+5} \right)\)  \(9{{x}^{2}}+25\) (prime – can’t factor with reals, would be \(\left( {3x+5i} \right)\left( {3x5i} \right)\) using imaginary numbers) 
Factoring with Sum and Difference of Cubes
For cubic binomials (sum or difference of cubes), we can factor:
\(\begin{array}{l}{{a}^{3}}+{{b}^{3}}=\left( {a+b} \right)\left( {{{a}^{2}}ab+{{b}^{2}}} \right)\\{{a}^{3}}{{b}^{3}}=\left( {ab} \right)\left( {{{a}^{2}}+ab+{{b}^{2}}} \right)\\\text{ signs: }S\text{ame }O\text{pp }\,A\text{lways }P\text{ositive}\end{array}\) 
Remember: S O A P when it comes to signs of cubic factoring. 
Here are some examples of factoring sums and differences of cubes:
Difference of Cubes 
Sum of Cubes 
Difference of Cubes 
\({{x}^{3}}8{{y}^{3}}\text{ }\)
\(\begin{align}(a&=\sqrt[3]{{{{x}^{3}}}}=x,\,\,\,\,\,\,b=\sqrt[3]{{8{{y}^{3}}}}=2y)\\\,\,\,\,\,{{x}^{3}}8{{y}^{3}}&=\left( {x2y} \right)\left[ {{{{\left( x \right)}}^{2}}+\left( x \right)\left( {2y} \right)+{{{\left( {2y} \right)}}^{2}}} \right]\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&=\left( {x2y} \right)\left( {{{x}^{2}}+2xy+4{{y}^{2}}} \right)\end{align}\) 
\(27{{y}^{3}}+1\)
\(\begin{align}(a&=\sqrt[3]{{27{{y}^{3}}}}=3y,\,\,\,\,\,\,b=\sqrt[3]{{{{1}^{3}}}}=1)\\&=\left( {3y+1} \right)\left[ {{{{\left( {3y} \right)}}^{2}}\left( {3y} \right)\left( 1 \right)+{{1}^{2}}} \right]\\&=\left( {3y+1} \right)\left( {9{{y}^{2}}3y+1} \right)\end{align}\) 
\(8{{x}^{3}}{{y}^{6}}125{{z}^{3}}\)
\(\begin{align}\,(a&=\sqrt[3]{{8{{x}^{3}}{{y}^{6}}}}=2x{{y}^{2}},\,\,\,\,\,\,b=\sqrt[3]{{125{{z}^{3}}}}=5z)\\&=\left( {2x{{y}^{2}}5z} \right)\left[ {{{{\left( {2x{{y}^{2}}} \right)}}^{2}}+\left( {2x{{y}^{2}}} \right)\left( {5z} \right)+{{{\left( {5z} \right)}}^{2}}} \right]\\&=\left( {2x{{y}^{2}}5z} \right)\left( {4{{x}^{2}}{{y}^{4}}+10x{{y}^{2}}z+25{{z}^{2}}} \right)\end{align}\)

Factoring with Polynomials
We first learned about factoring methods here in the Solving Quadratics by Factoring and Completing the Square section. We will learn how to solve Polynomials in the Graphing and Finding Roots of Polynomial Functions section in order to do more factoring, as shown here.
Remember that when we want to find solutions or roots, we set the equation to 0, factor, set each factor to 0 and solve.
Here are some examples of factoring and solving polynomial equations; solve over the reals:
Equation 
Solution(s) 
Explanation 
\(2{{x}^{3}}12{{x}^{2}}+18x=0\)  \(\begin{array}{c}2{{x}^{3}}12{{x}^{2}}+18x=0\\2x\left( {{{x}^{2}}6x+9} \right)=0\\2x{{\left( {x3} \right)}^{2}}=0\\2x=0;\,\,\,\,\,\,\,x3=0\\x=0,\,\,\,3\end{array}\) 
We first have to take the greatest common factor (GCF) out, which is \(2x\). We then have to factor (“unfoil”), but don’t forget to bring the “\(2x\)” down as part of the answer. (Remember when factoring that we need to find two numbers that add up to –6 but multiply to 9: –3 and –3.); this is actually a perfect square trinomial since when you double –3 we get –6, and when we square it, you get 9). Then set each factor to 0 to get the solutions or roots. 
\({{x}^{4}}+2{{x}^{3}}27x=54\)  \(\begin{array}{c}{{x}^{4}}+2{{x}^{3}}27x=54\\{{x}^{4}}+2{{x}^{3}}27x54=0\\{{x}^{3}}\left( {x+2} \right)27\left( {x+2} \right)=0\\\left( {{{x}^{3}}27} \right)\left( {x+2} \right)=0\\{{x}^{3}}27=0;\,\,\,x+2=0\\x=3,\,\,\,2\end{array}\) 
In order to be able to solve this polynomial, we must first set it to 0, so subtract 54 from both sides. Then separate the four terms into two sets of two terms (grouping) to see if we can see a pattern when we take out the GCF from each set of term. (We do: we are left with \(x+20\)!) Then, because of the distributive property, take out the expression that’s repeated, use that as one factor, and use the coefficients of this expression as the other factor. Then set each factor to 0 to get the solutions or roots. Note that we know that the cube root of 27 is 3. 
\(16{{x}^{4}}81=0\)  \(\displaystyle \begin{array}{c}16{{x}^{4}}81=0\\\left( {4{{x}^{4}}9} \right)\left( {4{{x}^{4}}+9} \right)=0\\\left( {2{{x}^{2}}3} \right)\left( {2{{x}^{2}}+3} \right)\left( {4{{x}^{4}}+9} \right)=0\\2{{x}^{2}}3=0\\x=\pm \frac{3}{2}\end{array}\) 
We can see that we have a difference of squares, and then a difference of squares again. Also remember that when we take the square root of a number, we have to consider both the positive and negative roots. Note that \(2{{x}^{2}}+3\) and \(4{{x}^{4}}+9\) (sum of squares) are prime, meaning we can’t factor them any further. Set the remaining factor to 0 to get the solutions or roots. 
\({{x}^{2}}2\sqrt{6}=6\)  \(\begin{array}{c}{{x}^{2}}2\sqrt{6}x=6\\{{x}^{2}}2\sqrt{6}x+6=0\\{{\left( {x\sqrt{6}} \right)}^{2}}=0\\x\sqrt{6}=0\\x=\sqrt{6}\end{array}\) 
First make sure we set the equation to 0 be adding 6 to both sides. This one’s a little tricky, since we have a square root sign. But notice that if we halve the coefficient of the middle term and then square it, we get the last term! So we actually have a perfect square trinomial. Set the factor to 0 and solve for \(x\). 
\({{x}^{3}}64=0\)  \(\begin{array}{c}\left( {x4} \right)\left( {{{x}^{2}}+4x+16} \right)=0\\x4=0\\x=4\end{array}\) 
There are actually two ways to do this; the easiest is to add 64 to both sides and then take the cube root of both sides to get \(x=4\). But let’s factor the difference of cubes for practice; remember that: \(\begin{array}{l}{{a}^{3}}+{{b}^{3}}=\left( {a+b} \right)\left( {{{a}^{2}}ab+{{b}^{2}}} \right)\\{{a}^{3}}{{b}^{3}}=\left( {ab} \right)\left( {{{a}^{2}}+ab+{{b}^{2}}} \right)\end{array}\). Since \({{x}^{2}}+4x+16\) is prime, we only have \(x=4\) (by setting the factor to 0). 
Here’s even more advanced solving, using techniques we will learn in the Graphing and Finding Roots of Polynomial Functions section.
Solve over the real and complex numbers:
Equation 
Solution(s) 
Explanation 
\(2{{x}^{3}}+3{{x}^{2}}11x6=0\)  \(\displaystyle \begin{array}{c}2{{x}^{3}}+3{{x}^{2}}11x6=0\\\left( {x+3} \right)\left( {2{{x}^{2}}3x2} \right)=0\\\left( {x+3} \right)\left( {2x+1} \right)\left( {x2} \right)=0\\x+3=0;\,\,\,\,\,2x+1=0;\,\,\,\,x2=0\\x=3,\,\,\frac{1}{2},\,\,2\end{array}\) 
We can either use a graphing calculator, or use guess and check to find a first root in this polynomial. We know from the Rational Root Test that possible factors are: \(\require{cancel} \displaystyle \frac{{\pm 1,\,\pm 2,\,\pm 3,\,\pm 6}}{{\pm 1,\,\pm 2}}\,=\,\pm \,\,1,\,\pm \,\,2,\,\pm \,\,3,\,\pm \,\,6,\,\pm \,\,\frac{1}{2},\cancel{{\pm \,\,1}},\pm \,\,\frac{3}{2},\cancel{{\pm \,\,3}}\) If we try some of these, we see that –3 works (so we know that is a factor). So now let’s use synthetic division to get the remaining factor(s): \begin{array}{l}\left. {\underline {\, Now we end up with \(2{{x}^{2}}3x2\). We can factor this to \(\left( {2x+1} \right)\left( {x2} \right)\). Then we set each factor to 0 to get the roots. 
\({{x}^{4}}+6{{x}^{3}}+6{{x}^{2}}+5x=0\)  \(\displaystyle \begin{array}{c}{{x}^{4}}+6{{x}^{3}}+6{{x}^{2}}+5x=0\\x\left( {{{x}^{3}}+6{{x}^{2}}+6x+5} \right)=0\\x\left( {x+5} \right)\left( {{{x}^{2}}+x+1} \right)=0\\x=0;\,\,\,x+5=0;\,\,\,{{x}^{2}}+x+1=0\\x=0,\,\,\,5,\,\,\,\frac{{1\pm \sqrt{3}\,i}}{2}\end{array}\) 
We can first factor out an \(x\), which is the GCF. Then we can “guess” that –5 is a root, using the Rational Root Test. So let’s use synthetic division to find the remaining factor: \begin{array}{l}\left. {\underline {\, We end up with \({{x}^{2}}+x+1\). It looks like we can’t factor, so let’s use Quadratic Formula, and we’ll end up with some complex roots:
\(\displaystyle \frac{{b\pm \sqrt{{{{b}^{2}}4ac}}}}{{2a}}=\frac{{1\pm \sqrt{{14\left( 1 \right)\left( 1 \right)}}}}{2}=\frac{{1\pm \sqrt{{3}}}}{2}=\frac{{1\pm \sqrt{3}\,i}}{2}\)
Set the other factors to 0, and we have all our factors. 
Factoring and Solving with Exponentials
Factoring and Solving with exponents can be a bit trickier. Note that we learned about the properties of exponents here in the Exponents and Radicals in Algebra Section, and did some solving with exponents here. Note also that we’ll discuss Exponential Functions here.
In your PreCalculus and Calculus classes, you may see algebraic exponential expressions that need factoring and possibly solving, either by taking out a Greatest Common Factor (GCF) or by “unfoiling”. These really aren’t that bad, if you remember a few hints:
 To take out a GFC with exponents, take out the factor with the smallest exponent, whether it’s positive or negative. (Remember that “larger” negative numbers are actually smaller). This is even true for fractional exponents. Then, to get what’s left in the parentheses after you take out the GCF, subtract the exponents from the one you took out. For example, for the expression \({{x}^{5}}+{{x}^{2}}+x\), we’d take out \({{x}^{5}}\) for the GCF to get \({{x}^{5}}\left( {{x}^{5\,5}}+{{x}^{2\,5}}+{{x}^{1\,5}} \right)={{x}^{5}}\left( {{x}^{0}}+{{x}^{3}}+{{x}^{6}} \right)={{x}^{5}}\left( 1+{{x}^{3}}+{{x}^{6}} \right)\). (Remember that “ – – ” is the same as “+ +” or “+”). Multiply back to make sure you’ve done it correctly.
 For fractional coefficients, find the common denominator, and take out the fraction that goes into all the other fractions. So for the fraction you take out, the denominator is the least common denominator of all the fractions, and the numerator is the Greatest Common Factor (GCF) of the numerators. For example, for the expression \(\frac{3}{4}{{x}^{2}}\frac{2}{4}x+\frac{16}{4}=\frac{1}{4}\left( 3{{x}^{2}}2x+16 \right)\) (since nothing except for 1 goes into 3 and 2 and 16). Multiply back to make sure you’ve done it correctly.
 For a trinomial with a constant, if the largest exponent is twice that of the middle exponent, then use substitution like u, for the middle exponent, “unfoil”, and then put the “real” expression back in. For example, for \({{x}^{\frac{2}{3}}}{{x}^{\frac{1}{3}}}2\) , let \(u={{x}^{\frac{1}{3}}}\) , and we have \({{u}^{2}}u2\) , which factors to \(\left( u2 \right)\left( u+1 \right)\). We can then translate back to \(\left( {{x}^{\frac{1}{3}}}2 \right)\left( {{x}^{\frac{1}{3}}}+1 \right)\) and solve from there (set each to 0 and solve). Always foil back to make sure you’ve done it correctly. We call this method usubstitution or simply usub.
Let’s do some factoring. Learning to factor these will actually help you a lot when you get to Calculus:
Math 
Notes 
\(\begin{array}{l}\color{#800000}{{4{{{\left( {x2} \right)}}^{{9}}}2{{{\left( {x2} \right)}}^{{10}}}}}\\=2{{\left( {x2} \right)}^{{10}}}\left[ {2{{{\left( {x2} \right)}}^{{9\left( {10} \right)}}}{{{\left( {x2} \right)}}^{{10\left( {10} \right)}}}} \right]\\=2{{\left( {x2} \right)}^{{10}}}\left[ {2{{{\left( {x2} \right)}}^{1}}{{{\left( {x2} \right)}}^{0}}} \right]\\=2{{\left( {x2} \right)}^{{10}}}\left( {2x41} \right)\\=2{{\left( {x2} \right)}^{{10}}}\left( {2x5} \right)\end{array}\) 
We need to take out the common factor (GCF) of the coefficients, and the smallest exponent of the algebraic terms: \(2{{\left( {x2} \right)}^{{10}}}\). Since the opposite of multiplication is division, and when we divide, we subtract exponents, starting from the exponent that will be inside the parentheses. (Try this with easier exponents to convince yourself!). When you subtract exponents, be sure to watch signs (remember that two negatives in a row is a positive). Also remember than anything raised to 0 is just 1. Then combine terms. It’s a good idea to multiply back before you combine terms to make sure you took out the GCF correctly! 
\(\begin{array}{l}\color{#800000}{{{{{\left( {{{x}^{2}}4} \right)}}^{{\frac{3}{5}}}}+5{{{\left( {{{x}^{2}}4} \right)}}^{{\frac{2}{5}}}}}}\\={{\left( {{{x}^{2}}4} \right)}^{{\frac{3}{5}}}}\left[ {{{{\left( {{{x}^{2}}4} \right)}}^{{\frac{3}{5}\left( {\frac{3}{5}} \right)}}}+5{{{\left( {{{x}^{2}}4} \right)}}^{{\frac{2}{5}\left( {\frac{3}{5}} \right)}}}} \right]\\={{\left( {{{x}^{2}}4} \right)}^{{\frac{3}{5}}}}\left[ {{{{\left( {{{x}^{2}}4} \right)}}^{0}}+5{{{\left( {{{x}^{2}}4} \right)}}^{1}}} \right]\\={{\left( {{{x}^{2}}4} \right)}^{{\frac{3}{5}}}}\left( {1+\,\,\,\,5{{x}^{2}}20} \right)\\={{\left( {{{x}^{2}}4} \right)}^{{\frac{3}{5}}}}\left( {5{{x}^{2}}19} \right)\end{array}\) 
We do this one the same way, but be careful with the fractional exponents. Again, take out the algebraic terms with the smallest exponent: \({{\left( {{{x}^{2}}4} \right)}^{{\frac{3}{5}}}}\).
Note that \(\displaystyle \frac{2}{5}\left( {\frac{3}{5}} \right)=\frac{5}{5}=1\). 
\(\displaystyle \begin{array}{l}\color{#800000}{{3{{x}^{{\frac{9}{2}}}}15{{x}^{{\frac{7}{2}}}}+12{{x}^{{\frac{5}{2}}}}}}\\=3{{x}^{{\frac{5}{2}}}}\left( {{{x}^{{\frac{9}{2}\frac{5}{2}}}}5{{x}^{{\frac{7}{2}\frac{5}{2}}}}+4{{x}^{{\frac{5}{2}\frac{5}{2}}}}} \right)\\=3{{x}^{{\frac{5}{2}}}}\left( {{{x}^{2}}5{{x}^{1}}+4{{x}^{0}}} \right)\\=3{{x}^{{\frac{5}{2}}}}\left( {{{x}^{2}}5x+4} \right)\,\,=\,\,3{{x}^{{\frac{5}{2}}}}\left( {x4} \right)\left( {x1} \right)\end{array}\) 
This is a trinomial that we might be able to “unfoil”, but we still need to take out any possible GCF first. We take out the GCF of the coefficients, and then the exponential term with the smallest exponent. Note then that we can “unfoil” after we take out the GCF. 
\(\begin{array}{l}\,\,\color{#800000}{{\,\frac{3}{8}{{{\left( {x5} \right)}}^{{\frac{3}{4}}}}+\frac{5}{4}{{{\left( {x5} \right)}}^{{\frac{1}{4}}}}}}\\=\frac{3}{8}{{\left( {x5} \right)}^{{\frac{3}{4}}}}+\frac{{10}}{8}{{\left( {x5} \right)}^{{\frac{1}{4}}}}\\=\frac{1}{8}{{\left( {x5} \right)}^{{\frac{3}{4}}}}\left[ {3{{{\left( {x5} \right)}}^{{\frac{3}{4}\left( {\frac{3}{4}} \right)}}}+10{{{\left( {x5} \right)}}^{{\frac{1}{4}\left( {\frac{3}{4}} \right)}}}} \right]\\=\frac{1}{8}{{\left( {x5} \right)}^{{\frac{3}{4}}}}\left[ {3{{{\left( {x5} \right)}}^{0}}+10{{{\left( {x5} \right)}}^{1}}} \right]\\=\frac{1}{8}{{\left( {x5} \right)}^{{\frac{3}{4}}}}\left[ {3+10\left( {x5} \right)} \right]\\=\frac{1}{8}{{\left( {x5} \right)}^{{\frac{3}{4}}}}\left[ {10x47} \right]\end{array}\) 
This one’s a little trickier with the fractions, but let’s use the rules we talked about earlier: Find the common denominator (8), and take out the fraction that goes into all the other fractions \(\displaystyle \left( {\frac{1}{8}} \right)\). So for the fraction you take out, the denominator is the LCD of all the fractions (8), and the numerator is the GCF of the numerators (only 1 goes into both 3 and 10). Notice that when you take the \(\displaystyle \frac{1}{8}\) out, you are just left with the numerators (3 and 10, respectively). Always multiply back (before combining terms) to make sure you took out the GCF correctly.

After factoring, you may be asked to solve the exponential equation. Here are some examples, some using usubstitution.
Note that the third problem uses log solving that we will learn here in the Logarithmic Functions section.
Math 
Notes 
\(\displaystyle 2{{x}^{{\frac{1}{2}}}}+{{x}^{{\frac{1}{4}}}}15=0\)
Let \(\displaystyle u={{x}^{{\frac{1}{4}}}}\) \(\displaystyle \begin{array}{c}2{{u}^{2}}+u15=0\,\,\,\,\,\,\,\,\,\,\left( {2u5} \right)\left( {u+3} \right)=0\\u=\frac{5}{2},3\end{array}\)
Substitute \(\displaystyle {{x}^{{\frac{1}{4}}}}\) back in for \(u\) for both factors: \(\displaystyle {{x}^{{\frac{1}{4}}}}=\frac{5}{2};\,\,\,{{\left( {{{x}^{{\frac{1}{4}}}}} \right)}^{4}}={{\left( {\frac{5}{2}} \right)}^{4}};\,\,\,x=\frac{{625}}{{16}}\) \(\require{cancel} \displaystyle \cancel{{{{x}^{{\frac{1}{4}}}}=3}}\); No Solution
Check: 
Let’s use \(u\)substitution here, since we see that the largest exponent \(\left( {\frac{1}{2}} \right)\) is twice that of the middle exponent \(\left( {\frac{1}{4}} \right)\). Use the trick to let \(u\) equal to the variable in the middle: \(u={{x}^{{\frac{1}{4}}}}\), and then \(\displaystyle 2{{x}^{{\frac{1}{2}}}}=2{{\left( {{{x}^{{\frac{1}{4}}}}} \right)}^{2}}=2{{u}^{2}}\). (Remember that when we raise an exponent to another exponent, we multiply exponents).
Now we can factor more easily. We can factor this one; if you can’t, you can always use the Quadratic Formula.
Then we’ll put the \(u\)’s back in. Remember that we want to end up with \(\displaystyle {{x}^{1}}\) by itself, so we have to raise both sides to the reciprocal of the exponent. For the second factor, we have no solution, since we have an even exponent in the denominator and we need to end up with a negative number – can’t be done with real numbers.
Try the answer back in for \(x\) to make sure it works; \(\frac{{625}}{{16}}\) works! You can use the STO> X function in your graphing calculator to check (see left). 
\(\displaystyle {{9}^{x}}+{{3}^{{x+1}}}=4\) \(\displaystyle \begin{array}{l}{{\left( {{{3}^{2}}} \right)}^{x}}+\left( {{{3}^{x}}} \right)\left( {{{3}^{1}}} \right)=4\\{{3}^{{2x}}}+\left( 3 \right){{3}^{x}}=4\\{{\left( {{{3}^{x}}} \right)}^{2}}+\left( 3 \right){{3}^{x}}=4\end{array}\)
Let \(\displaystyle u={{3}^{x}}\) \(\begin{array}{c}{{u}^{2}}+3u4=0\,\,\,\,\,\,\,\,\,\,\left( {u1} \right)\left( {u+4} \right)=0\\u=1,\,4\end{array}\)
Substitute \({{3}^{x}}\) back in for \(u\) for both factors: \({{3}^{x}}=1;\,\,\,\,\,x=0\) \(\cancel{{{{3}^{x}}=4}}\); No Solution 
It looks like we need the “3”’s to match, so let’s break up the second term and change the base of the first term – it works! Remember that when we multiply the same base, we add exponents. Also remember that when we raise an exponent to another exponent, we multiply exponents.
We’ll then let \(u={{3}^{x}}\), and then \({{9}^{x}}={{\left( {{{3}^{2}}} \right)}^{x}}={{3}^{{2x}}}={{3}^{{x\cdot 2}}}={{\left( {{{3}^{x}}} \right)}^{2}}={{u}^{2}}\).
For the first factor, we just know that anything raised to 0 = 1, so the solution is 0. For the second factor, we get no solution, since we can’t raise 3 to anything to get a negative number.
Try the 0 back in the original and see if it works – it does: \({{9}^{0}}+\left( 3 \right){{3}^{0}}=4;\,\,\,\,1+3\left( 1 \right)=4\) 
\({{\left( {{{3}^{x}}27} \right)}^{3}}+18{{\left( {{{3}^{x}}27} \right)}^{2}}=0\) \(\displaystyle \begin{array}{c}{{\left( {{{3}^{x}}27} \right)}^{2}}\left[ {{{{\left( {{{3}^{x}}27} \right)}}^{1}}+18} \right]=0\\{{\left( {{{3}^{x}}27} \right)}^{2}}\left( {{{3}^{x}}9} \right)=0\\\sqrt{{{{{\left( {{{3}^{x}}27} \right)}}^{2}}}}=\pm 0=0;\,\,\,\,\,\,\,\left( {{{3}^{x}}9} \right)=0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{3}^{x}}=27;\,\,\,\,\,\,\,\,\,{{3}^{x}}=9\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=3;\,\,\,\,\,\,\,\,\,x=2\end{array}\) 
For this one, we don’t really need \(u\)sub, but since we can just take out the expression with the smallest exponent, like we did earlier.
After factoring, we set each factor to 0. We have to think about what exponents will work when 3 is raised to that exponent – to get 27 and 9 (or we could have used logs).
We get that both 3 and 2 work; plug them back in the original to make sure they are correct. 
Here’s one where we have to use logs to solve:
\(2{{e}^{{5x}}}5{{e}^{{2x}}}12{{e}^{{x}}}=0\) \(\displaystyle \begin{array}{c}{{e}^{{x}}}\left( {2{{e}^{{6x}}}5{{e}^{{3x}}}12} \right)=0\\\\\text{Let }u={{e}^{{3x}}}\\{{e}^{{x}}}\left( {2{{u}^{2}}5u12} \right)=0\\{{e}^{{x}}}\left( {2u+3} \right)\left( {u4} \right)=0\\{{e}^{{x}}}=0\,\,\,\,\,\,\,\,\,u=\frac{3}{2}\,\,\,\,\,\,\,\,\,\,u=4\\{{e}^{{x}}}=0\,\,\,\,\,\,\,\,\,{{e}^{{3x}}}=\frac{3}{2}\,\,\,\,\,\,\,\,\,\,{{e}^{{3x}}}=4\\\,\,\,\,\cancel{{{{e}^{{x}}}=0}}\,\,\,\,\,\,\,\,\,\cancel{{{{e}^{{3x}}}=\frac{3}{2}}}\,\,\,\,\,\,\,\,\ln {{e}^{{3x}}}=\ln 4\\3x=\ln 4;\,\,\,\,\,\,\,\,x=\frac{{\ln 4}}{3}\end{array}\) 
First, factor out the GCF, which is \({{3}^{{x}}}\) (smallest exponent).
Then, set \(u={{e}^{{3x}}}\), since \(\displaystyle {{e}^{{6x}}}={{\left( {{{e}^{{3x}}}} \right)}^{2}}\) (note that \(u\) is in the middle term).
Factor, and set each factor to 0 (including the GCF).
Note that we had to “throw away” the extraneous solutions, since (positive) \(e\) raised to anything can’t be 0 or negative. 
Learn these rules, and practice, practice, practice!
For Practice: Use the Mathway widget below to try a Factoring problem. Click on Submit (the blue arrow to the right of the problem) and click on Factor to see the answer.
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On to Rational Functions and Equations – you are ready!